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\begin{document}
\title{Analitik Geometri \"Ozeti}
\subtitle{Analytic Geometry Summary\\
with English foreward}
\author{David Pierce}
\date{14 May\i s 2015\\
G\"ozden ge\c cirilmi\c s 8 Nisan 2016}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}
\maketitle

%\input{analitik-preface} % the file analitik-geometri-ozeti-2.* on the course webpage does not have this English preface.  Now I just include it:

\chapter*{\.Ingilizce \"ons\"oz% // \emph{English preface}
}

\selectlanguage{english}
These notes are a summary of a course in analytic geometry
given to first-year students
in the mathematics department of Mimar Sinan Fine Arts University,
Istanbul, in the second---the spring---semester of 2014--5.

The notes are divided into paragraphs, which are numbered serially.
Each paragraph is further labelled as a definition, a remark, or a theorem.
A few theorems are given explicit proofs.
Most proofs are omitted.
These proofs may have been given in class,
or they may have been left as exercises.
Other exercises were given in class, but these are not included here.

The course aims to avoid two failings 
of contemporary presentations of analytic geometry:%
\nocite{NFB}\nocite{US-AG}
\begin{inparaenum}[(1)]
\item 
the logical foundations of the subject are not clear, and
\item
conic sections are not explained as such.
\end{inparaenum}
The first is a failure of honesty:
honesty to our students and ourselves
about what we assume, and what we can actually prove.
The second is a failure to share the 
liberating power of mathematics,
replacing it with rote memorization.
If students are going to learn the parabola, hyperbola, and ellipse,
they ought to know what the names really mean.

\section*{Foundations}

When Descartes invented analytic geometry,
he had Ancient Greek mathematics as a foundation.
He did \emph{not} draw two perpendicular lines in a plane
and use them to establish a one-to-one correspondence 
between points in the plane and ordered pairs of numbers.
He \emph{did} introduce the practice, which we follow today,
of naming lengths by minuscule Latin letters:
known lengths from the beginning of the alphabet,
and unknown lengths from the end.
By introducing a unit length,
Descartes showed that lengths could be manipulated algebraically---%
they could be added and multiplied---%
with full geometric justification.

Students today will come to a course in analytic geometry
with some knowledge of the ordered field of real numbers.
They can conceive this ordered field as an unbounded straight line.
An analytic geometry textbook will use pairs of real numbers 
to coordinatize a plane in the manner referred to above.
Then the book will give a rule for expressing the distance 
between two points of the plane so coordinatized.
The rule will be justified by a vague reference to the Pythagorean Theorem.

This justification is inadequate,
if what is meant by the Pythagorean Theorem
is Proposition 47 of Book \textsc i of Euclid's \emph{Elements.}
This proposition is that the square on the hypotenuse of a right triangle
is equal to the squares on the two legs.
This means the two small squares can be cut into pieces
and rearranged to form the large square.
This process has no obvious connection
to adding the so-called ``squares'' of two real numbers.
We can make the connection in two ways.
\begin{compactenum}
  \item
We can develop the set of ordered pairs of real numbers 
into an inner product space.
Here we \emph{define} the notions of length and angle,
and we prove that they have the properties that we want.
\item
Alternatively, we can use geometry to give an infinite straight line
the structure of an ordered field.
\end{compactenum}
Following Descartes, we take the latter approach in the present course.
The students have already had a course in which they themselves demonstrate, 
at the board, 
the contents of Book \textsc i of Euclid's \emph{Elements.}
That book then is the logical foundation of the notes below.

We need a theory of proportion, so that,
after choosing a unit length, 
we can define the product $a\cdot b$ of two lengths by
\begin{equation*}
  1:b::a:a\cdot b.
\end{equation*}
Descartes presumably relies on the theory developed in Books \textsc v
and \textsc{vi} of the \emph{Elements.}
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-4.6,-0.5)(0,2.7)
    \psline(-4.6,0)(0,0)(-3.6,2.8)
\psline(-3,0)(-2.7,2.1)
\psline(-2,0)(-1.8,1.4)
\uput[d](-2,0){$A$}
\uput[d](0,0){$B$}
\uput[ur](-1.8,1.4){$C$}
\uput[d](-3,0){$D$}
\uput[ur](-2.7,2.1){$E$}
  \end{pspicture}

  
\end{figure}
By this theory, in the figure, assuming $DE\parallel AC$, then
\begin{equation}\label{eqn:prop-def}
  BA:BD::BC:BE.
\end{equation}
In particular, if $AB=1$, $BC=a$, and $BD=b$,
then we define $BE=a\cdot b$.

Euclid's theory of proportion relies on the so-called Archi\-medean Axiom:
that of any two lengths, some multiple of either exceeds the other.
(Moreover, 
some multiple of the \emph{difference} of the lengths exceeds the greater.)
David Hilbert develops a theory of proportion
without using this axiom.
First he defines multiplication as Descartes does;
but he must assume that angle $ABC$ is right.
He proves commutativity and associativity of multiplication
by means of what he calls Pascal's Theorem,
though it was known to Pappus of Alexandria.%
\nocite{MR654680}\nocite{MR0472307}\nocite{Pappus}
Then he defines
\begin{equation}\label{eqn:mult}
  a:b::c:d\iff ad=bc.
\end{equation}
I have preferred to proceed more historically,
though without using the Archimedean Axiom.
I make \eqref{eqn:prop-def} the definition of proportion,
assuming angle $ABC$ in the figure is right.
It takes some work to show that the same proportion holds,
regardless of the angle.
Instead of Pascal's or Pappus's Theorem,
our main tool is Proposition \textsc i.43 of the \emph{Elements}
and its converse
(here incorporated into Theorem \ref{thm:I.43} below).
This gives us \eqref{eqn:mult},
but on the understanding that the product of two lengths is an \emph{area.}
The product of three lengths is likewise a \emph{volume.}
Ultimately, by defining a unit length like Descartes,
we can represent areas and volumes by lengths;
but there seems to be no need to do this
until we do coordinatize the plane,
and we want to use an equation like
\begin{equation*}
  y=mx+b
\end{equation*}
to define a straight line.
(Even then, without defining a unit, 
we could say $m$ is a ratio, not a length.)

We make explicit that a \emph{length}
is a congruence-class of line segments.
For Euclid, our line segments are straight lines,
and congruence of them is equality.
In general, the measure of a magnitude
can be understood as its equality-class.
We do not follow Archimedes in assuming that the circumference of a circle 
has a ratio to its diameter.
For us, $\uppi$ is just the double of the measure of a right angle.

\section*{Conic sections}

Many textbooks today \emph{tell} you that the parabola, hyperbola, and ellipse
can be obtained by cutting a right cone with a plane.
But they \emph{define} each of these curves by means of an equation,
or else they derive the equation from a focus and directrix,
and they usually do not bother to show you 
that the property expressed by the equation
can actually be \emph{proved} for the appropriate conic section.
Hilbert and Cohn-Vossen prove it beautifully,\nocite{MR0046650}
but using \emph{right} cones only.
Apollonius proves it for arbitrary cones,
and I take his approach here.
I look at the conic sections as soon as possible,
because they are beautiful in themselves,
and because they show the power of analytic geometry
to encode beauty in equations.
My approach then asks the student to think in three dimensions
near the beginning of the course.
But there is reason for it:
to see how a third dimension illuminates two.

\selectlanguage{turkish}


%\begin{multicols}2
\tableofcontents
%  \end{multicols}
\listoffigures

\chapter{Orant\i lar}

\section{Denklik ba\u g\i nt\i lar\i}

\begin{definition}
\textbf{Do\u gal say\i lar,} $1$, $2$, $3$, .\lips Bunlar 
$\N$ k\"umesini olu\c sturur:
  \begin{equation*}
  \N=\{1,2,3,\dots\}.
  \end{equation*}
(Bu ifadede ``$=$'' i\c sareti \emph{ayn\i l\i\u g\i} g\"osterir,
yani $\N$ ve $\{1,2,3,\dots\}$ ayn\i\ k\"umedir.)
\end{definition}

\begin{remark}
\.Ilkokuldan bildi\u gimiz gibi 
iki do\u gal say\i\ toplanabilir ve \c carp\i labilir,
ve do\u gal say\i lar s\i ralan\i r.
\end{remark}

\begin{definition}
S\i ral\i\ ikililer,
\begin{equation*}
  (a,b)=(x,y)\iff a=x\And b=y
\end{equation*}
\"ozelli\u gini sa\u glar.
T\"um $A$ ve $B$ k\"umeleri i\c cin
\begin{equation*}
A\times B=\{(x,y)\colon x\in A\And y\in B\}.
\end{equation*}
\end{definition}



\begin{definition}
Bir $A$ k\"umesinin yans\i mal\i, simetrik, ve ge\c ci\c sli 
$2$-konumlu $R$ ba\u g\i nt\i s\i,
$A$ k\"umesinin \textbf{denklik ba\u g\i nt\i s\i d\i r.}
$A$ k\"umesinin $b$ eleman\i n\i n
\textbf{$R$ ba\u g\i nt\i s\i na g\"ore denklik s\i n\i f\i}
veya \textbf{$R$-s\i n\i f\i,}
\begin{equation*}
  \{x\in A\colon x\mathrel Rb\}
\end{equation*}
k\"umesidir.%%%%%
\footnote{Bu k\"umeye ``denklik s\i n\i f\i'' demek, bir gelenektir.
K\"umeler kuram\i nda her k\"ume bir s\i n\i ft\i r,
ama her s\i n\i f k\"ume de\u gildir.
\"Orne\u gin $\{x\colon x\notin x\}$ s\i n\i f\i, k\"ume olamaz.}
%%%%%%%%%%%%%%%%
Bu denklik s\i n\i f\i\ i\c cin
\begin{equation*}
  [b]
\end{equation*}
k\i saltmas\i\ kullan\i labilir
(ama \Teoremde{thm:eq-cl}n sonra kullan\i lmayacak).
\end{definition}

\begin{theorem}\label{thm:eq-cl}
  $R$, $A$ k\"umesinin denklik ba\u g\i nt\i s\i\ olsun, 
ve $b\in A$, $c\in A$ olsun.
O zaman
\begin{equation*}
  \qquad[b]=[c]\qquad\text{ veya }\qquad[b]\cap[c]=\emptyset.
\end{equation*}
%  Ya
%\begin{equation*}
%[b]=[c]
%\end{equation*}
%ya da
%\begin{equation*}
%[b]\cap[c]=\emptyset.
%\end{equation*}
\end{theorem}

\begin{definition}
  $\N\times\N$ k\"umesinin $\approx$ ba\u g\i nt\i s\i,
  \begin{equation*}
    (k,\ell)\approx(x,y)\iff ky=\ell x
  \end{equation*}
tan\i m\i n\i\ sa\u glas\i n.
\end{definition}

\begin{theorem}
  $\N\times\N$ k\"umesinin $\approx$ ba\u g\i nt\i s\i,
denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{definition}
$\N\times\N$ k\"umesinin $(k,\ell)$ eleman\i n\i n $\approx$-s\i n\i f\i,
  \begin{equation*}
    \frac k{\ell}
  \end{equation*}
veya $k/\ell$ \textbf{pozitif kesirli say\i s\i d\i r.}
Pozitif kesirli say\i lar,
\begin{equation*}
  \Qp
\end{equation*}
k\"umesini olu\c sturur.
\end{definition}

\begin{theorem}
  A\c sa\u g\i daki e\c sitlikler,
$\Qp$ k\"umesinin toplama ve \c carpma i\c slemleri i\c cin iyi tan\i md\i r:
  \begin{align*}
    \frac k\ell+\frac mn&=\frac{kn+\ell m}{\ell n},&\frac k\ell\cdot\frac mn&=\frac{am}{\ell n}.
  \end{align*}
Yani
\begin{multline*}
  \frac k\ell=\frac{k'}{\ell'}\And\frac mn=\frac{m'}{n'}\\
\implies
\frac{kn+\ell m}{\ell n}=\frac{k'n'+\ell'm'}{\ell'n'}\And
\frac{km}{\ell n}=\frac{k'm'}{\ell'n'}.
\end{multline*}
Ayr\i ca $\Qp$ a\c sa\u g\i daki tan\i ma g\"ore s\i ralan\i r:
\begin{equation*}
  \frac k\ell<\frac mn\iff kn<\ell m.
\end{equation*}
\end{theorem}

\section{Uzunluklar}

\begin{definition}
  \"Oklid'deki gibi\nocite{Oklid-2014-T},
normalde bir \textbf{do\u grunun} u\c c noktalar\i\ vard\i r.
\"Oklid'in 4.\ Ortak Kavram\i ndaki gibi
\c cak\i\c san do\u grular \textbf{e\c sittir.}
(\"Ozel olarak do\u grular i\c cin e\c sitlik ayn\i l\i k de\u gildir.)
\end{definition}

\begin{theorem}
  Do\u grular\i n e\c sitli\u gi, bir denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{proof}
  \"Oklid'in 1.\ Ortak Kavram\i na g\"ore e\c sitlik ge\c ci\c slidir.
\c Cak\i\c sman\i n yans\i mal\i l\i\u g\i\ ve simetrisi,
a\c c\i k olarak say\i labilir.
\end{proof}

\begin{definition}
Bir do\u grunun e\c sitlik s\i n\i f\i, do\u grunun \textbf{uzunlu\u gudur.}
K\"u\c c\"uk $a$, $b$, $c$, \lips Latin harfleri uzunluk g\"osterecek.
E\u ger bir $AB$ do\u grusunun uzunlu\u gu $c$ ise
\begin{equation*}
  AB=c
\end{equation*}
ifadesini yazar\i z.%%%%%
\footnote{Bu uygulama Descartes'\i n 1637 \emph{Geometri} kitab\i ndan gelir.}
\end{definition}

\begin{theorem}
\.Iki uzunluk toplanabilir, 
ve bir kesirli say\i\ bir uzunlu\u gu \c co\u galtabilir.
Toplama de\u gi\c smeli ve birle\c smelidir,
ve \c co\u galtma toplama \"uzerine da\u g\i l\i r.
E\u ger $a<b$ ise
\begin{equation*}
  a+x=b
\end{equation*}
denklemi \c c\"oz\"ulebilir.
\end{theorem}

\begin{definition}
$\mathscr R$, a\c sa\u g\i daki gibi tan\i mlanan ba\u g\i nt\i\ olsun.
$AB$, $CD$, $EF$, ve $GH$ do\u grular\i\ verilmi\c s olsun.
Baz\i\ $K$ ve $L$ noktalar\i\ i\c cin,
e\u ger $CD=BK$ ve $GH=FL$ ise,
ve \Sekilde{fig:ABCDEF}ki gibi
\begin{figure}[ht]
\mbox{}\hfill
\psset{unit=4mm}
  \begin{pspicture}(-3,-0.83)(4,3)
    \pspolygon(0,0)(4,3)(4,0)
\uput[d](0,0){$A$}
\uput[d](4,0){$B$}
\uput[u](4,3){$K$}
\psline(-3,3)(0,3)
\uput[u](-3,3){$C$}
\uput[u](0,3){$D$}
%\uput[ur](2,1.5){$e$}
%\uput[l](0,1.5){$a$}
%\uput[d](2,0){$b$}
  \end{pspicture}
\hfill\psset{unit=5mm}
  \begin{pspicture}(-3,-0.5)(4,3.5)
    \pspolygon(0,0)(4,3)(4,0)
\uput[d](0,0){$E$}
\uput[d](4,0){$F$}
\uput[u](4,3){$L$}
\psline(-3,3)(0,3)
\uput[u](-3,3){$G$}
\uput[u](0,3){$H$}
%\uput[ur](2,1.5){$f$}
%\uput[l](0,1.5){$c$}
%\uput[d](2,0){$d$}
  \end{pspicture}
\hfill\mbox{}
  \caption{Bir ba\u g\i nt\i}\label{fig:ABCDEF}
  
\end{figure}
$ABK$ ve $EFL$ \"u\c cgenlerinde $\angle ABK$ ve $\angle EFL$ dik
ve $\angle BAK=\angle FEL$ ise,
o zaman
\begin{equation*}
  (AB,CD)\mathrel{\mathscr R}(EF,GH)
\end{equation*}
olsun.
\end{definition}

\begin{theorem}\label{thm:prop}
$\mathscr R$ ba\u g\i nt\i s\i,
bir denklik ba\u g\i nt\i s\i d\i r.
Ayr\i ca $\mathscr R$
sadece do\u grular\i n uzunlu\u guna ba\u gl\i d\i r.
\end{theorem}

\begin{definition}\sloppy
E\u ger \Teoremde{thm:prop}ki gibi 
$(AB,CD)\mathrel{\mathscr R}(EF,GH)$ ise
$AB$, $CD$, $EF$, ve $GH$ do\u grular\i\ \textbf{orant\i l\i d\i r,}
ve
\begin{equation*}
  AB:CD::EF:GH
\end{equation*}
\textbf{orant\i s\i n\i} yazar\i z; ayr\i ca
$AB=a$, $CD=b$, $EF=c$, ve $GH=d$ ise
\begin{equation*}
  a:b::c:d
\end{equation*}
ifadesini yazar\i z.
Buradaki $AB:CD$ ve $a:b$ ifadeleri,
$(AB,CD)$ ve $(a,b)$ 
s\i ral\i\ ikililerinin denklik s\i n\i f\i n\i\ g\"osterir;
bu s\i n\i f, bir \textbf{orand\i r.}
Bu durumda ``$::$'' simgesi, 
oranlar\i n \emph{ayn\i l\i\u g\i n\i} g\"osterir.
(Bundan sonra $\mathscr R$ kullan\i lmayacak.)
\end{definition}

\begin{remark}
\c Simdi $a:b::c:d$ orant\i s\i,
\Sekilde{fig:prop}ki gibi g\"osterilebilir.
  \begin{figure}[ht]
\centering
    \psset{unit=4mm,labelsep=3pt}
  \begin{pspicture}(0,-1)(4,3)
    \pspolygon(0,0)(4,0)(4,3)
\psline(3.5,0)(3.5,0.5)(4,0.5)
\uput[d](2,0){$b$}
\uput[r](4,1.5){$a$}
\psarc(0,0){1}0{36.86}
%\uput{12pt}[18.4](0,0){$\theta$}
  \end{pspicture}
\qquad
\psset{unit=5mm}
  \begin{pspicture}(0,-0.8)(4,3)
    \pspolygon(0,0)(4,0)(4,3)
\psline(3.6,0)(3.6,0.4)(4,0.4)
\uput[d](2,0){$d$}
\uput[r](4,1.5){$c$}
\psarc(0,0){0.8}0{36.86}
%\uput{12pt}[18.4](0,0){$\theta$}
  \end{pspicture}

    \caption{Orant\i l\i l\i k}\label{fig:prop}
    
  \end{figure}
\end{remark}

\begin{theorem}
  \Sekilde{fig:parallel}
\begin{figure}[ht]
    \centering
\psset{unit=3mm}
    \begin{pspicture}(0,-1)(12,10)
      \pspolygon(0,0)(12,0)(12,9)
\psline(4,0)(12,6)
%\psline[linestyle=dotted](6,0)(12,6)
\uput[d](0,0){$A$}
\uput[d](12,0){$B$}
\uput[u](12,9){$C$}
\uput[d](4,0){$D$}
\uput[r](12,6){$E$}
%\uput{3pt}[100](6,0){$F$}
%\uput[r](12,3){$a$}
%\uput[d](8,0){$b$}
%\uput[u](9,0){$c$}
%\uput[r](12,7.5){$d$}
%\uput[d](2,0){$e$}
    \end{pspicture}
    \caption{Paralellik ve orant\i l\i l\i k}\label{fig:parallel}
    
  \end{figure}
$ABC$ a\c c\i s\i\ dik ise
  \begin{equation*}
    AB:BC::DB:BE\iff AC\parallel DE.
  \end{equation*}
\end{theorem}

\begin{theorem}
  $a:b::a:c\implies b=c$.
\end{theorem}

\begin{theorem}
 $a:b::c:d\implies a:b::a\pm c:b\pm d$.
(\Sekle{fig:sum} bak\i n.)
  \begin{figure}
    \centering\psset{unit=2mm,labelsep=1pt}
    \begin{pspicture}(20,15)
      \pspolygon(0,0)(20,0)(20,15)
\psline(8,6)(8,0)(20,9)
\uput[l](8,3){$a$}
\uput[u](4,0){$b$}
\uput[l](20,4.5){$c$}
\uput[u](14,0){$d$}
%\uput[l](20,12){$a$}
    \end{pspicture}
    \caption{Toplama}\label{fig:sum}
    
  \end{figure}
\end{theorem}

\begin{definition}
$a:c::c:b$ ise $c$ uzunlu\u guna $a$ ve $b$ uzunluklar\i n\i n
\textbf{orta orant\i l\i s\i} denir.
\end{definition}

\begin{theorem}\label{thm:axxb}
Her iki uzunlu\u gun orta orant\i l\i s\i\ vard\i r,
yani her
\begin{equation*}
a:x::x:b
\end{equation*}
orant\i s\i\ \c c\"oz\"ulebilir.
(\Sekle{fig:mean} bak\i n.)
  \begin{figure}
    \centering
\psset{unit=1.6mm,labelsep=2pt}
  \begin{pspicture}(25,12.5)
    \psline(0,0)(25,0)
\psline(9,0)(9,12)
\uput[u](4.5,0){$a$}
\uput[r](9,6){$x$}
\uput[u](17,0){$b$}
\psarc(12.5,0){12.5}{0}{180}
  \end{pspicture}
  \caption{Orta orant\i l\i}\label{fig:mean}
    
  \end{figure}
\end{theorem}

%\begin{theorem}
%  $a:b::d:e\And b:c::e:f\implies a:c::d:f$.
%\end{theorem}

\section{Alanlar}

\begin{theorem}\label{thm:I.43}
Ayn\i\ geni\c sli\u gi ve y\"uksekli\u gi olan dikd\"ortgenler e\c sittir.
\Sekilde{fig:ab=cx}
\begin{figure}[ht]
  \centering
\psset{unit=2mm}
  \begin{pspicture}(0,-1.5)(21,16.5)
    \pspolygon(0,0)(21,0)(21,14)(0,14)
\psline(0,10)(21,10)
\psline(6,0)(6,14)
\psline[linestyle=dotted](0,14)(21,0)
\uput[dl](0,0){$A$}
\uput[d](6,0){$B$}
\uput[dl](6,10){$C$}
\uput[l](0,10){$D$}
\uput[r](21,10){$E$}
\uput[dr](21,0){$F$}
\uput[ur](21,14){$H$}
\uput[ul](0,14){$G$}
\uput[u](6,14){$K$}
  \end{pspicture}
  \caption{Dikd\"ortgenlerin e\c sitli\u gi}\label{fig:ab=cx}
  
\end{figure}
$ABCD$ ve $CEHK$ dikd\"ortgenleri e\c sittir ancak ve ancak $GC$ ve $CF$
bir do\u grudad\i r.
\end{theorem}

\begin{definition}
Bir dikd\"ortgenin \textbf{alan\i,} onun e\c sitlik s\i n\i f\i d\i r.
Geni\c sli\u gi $a$ ve y\"uksekli\u gi $b$ olan dikd\"ortgenin alan\i
\begin{equation*}
  a\cdot b
\end{equation*}
veya $ab$ ile g\"osterilir.  Ayr\i ca $a\cdot a$ alan\i
\begin{equation*}
  a^2
\end{equation*}
ile g\"osterilir.
\end{definition}

\begin{theorem}
  Uzunluklar\i n \c carpmas\i\ de\u gi\c smelidir
ve toplama \"uzerine da\u g\i l\i r.
Ayr\i ca
\begin{equation*}
  ab=ac\implies b=c.
\end{equation*}
\end{theorem}

\begin{theorem}
  $a:b::c:d\iff ad=bc$.
\end{theorem}

\begin{theorem}
  $a:b::c:d\implies a:c::b:d$.
\end{theorem}

\begin{theorem}\label{thm:well}
  $ab=de\And ac=df\implies b:c::e:f$.
\end{theorem}

\begin{theorem}
  $a:b::d:e\And b:c::e:f\implies a:c::d:f$.
\end{theorem}

\begin{definition}
$c:d::b:e$ ise
  \begin{equation*}
    \comp abcd::a:e,
  \end{equation*}
ve $a:e$ oran\i, $a:b$ ve $c:d$ oranlar\i n\i n \textbf{bile\c skesidir.}
\end{definition}

\begin{theorem}
$a:b::c:d$ ve $e:f::g:h$ ise
  \begin{equation*}
  \comp abef::\comp cdgh.
  \end{equation*}
\end{theorem}

\begin{theorem}
  $\comp abcd::ac:bd$.
\end{theorem}

\begin{theorem}
  $\comp abcd::\comp cdab$.
\end{theorem}

\begin{theorem}\label{thm:sol}
Her $ab=cx$ denklemi \c c\"oz\"ulebilir.
\end{theorem}

\begin{definition}
  $a:b::c:d$ ise $d$ uzunlu\u guna $a$, $b$, ve $c$ uzunluklar\i n\i n
\textbf{d\"ord\"unc\"u orant\i l\i s\i} denir.
\end{definition}

\begin{theorem}
  Her \"u\c c uzunlu\u gun d\"ord\"unc\"u orant\i l\i s\i\ vard\i r, yani her
  \begin{equation*}
    a:b::c:x
  \end{equation*}
orant\i s\i\ \c c\"oz\"ulebilir.
\end{theorem}

\begin{definition}
$ab$ ve $cd$ alanlar\i\ verilmi\c s ise
\Teoreme{thm:sol} g\"ore bir $e$ i\c cin
%\begin{equation*}
  $cd=ae$.
%\end{equation*}
Bu durumda, tan\i ma g\"ore
\begin{equation*}
  ab:cd::a:e.
\end{equation*}
\Teorem{thm:well} sayesinde bu tan\i m iyidir,
yani $ab=fg$ ve $cd=hk$ ise $ab:cd::fg:hk$.
\end{definition}

\begin{theorem}
$ab:cd::ab:ef\implies cd=ef$.
\end{theorem}

\begin{theorem}
  $ab:cd::ef:gh\implies ab:ef::cd:gh$.
\end{theorem}

\begin{theorem}
  $ab:cd::ef:gh\implies ab:cd::ab+ef:cd+gh$.
\end{theorem}

\begin{theorem}
$a:b::c:d\iff a^2:b^2::c^2:d^2$.  
\end{theorem}

\begin{definition}
  A\c c\i lar\i\ s\i ras\i yla e\c sit olan \"u\c cgenler \textbf{benzerdir.}
\end{definition}

\begin{theorem}\sloppy
  Benzer \"u\c cgenlerin kenarlar\i\ orant\i l\i d\i r,
yani $ABC$ ve $DEF$ benzer ise
\begin{equation*}
  AB:BC::DE:EF.
\end{equation*}
\end{theorem}


\begin{theorem}
  \Sekilde{fig:parallel2}
  \begin{figure}[ht]
    \centering
\psset{unit=3mm}
    \begin{pspicture}(0,-1)(18,10)
%\psgrid
      \pspolygon(0,0)(12,0)(18,9)
\psline(4,0)(16,6)
%\psline[linestyle=dotted](6,0)(12,6)
\uput[d](0,0){$A$}
\uput[d](12,0){$B$}
\uput[u](18,9){$C$}
\uput[d](4,0){$D$}
\uput[dr](16,6){$E$}
    \end{pspicture}
    \caption{Paralellik ve orant\i l\i l\i k}\label{fig:parallel2}
    
  \end{figure}
$ABC$ herhangi \"u\c cgen olsun.  O zaman
  \begin{equation*}
    AB:BC::DB:BE\iff AC\parallel DE.
  \end{equation*}
\end{theorem}

\chapter{Koni kesitleri}

\section{Paraboller}

\begin{definition}\label{def:axis}
Bir daire
ve ayn\i\ d\"uzlemde olmayan bir nokta,
bir \textbf{koniyi} 
(\gr{k~wnos} ``\c cam kozala\u g\i'')
belirtir.
Daire, koninin \textbf{taban\i d\i r,}
ve nokta, koninin \textbf{tepe noktas\i d\i r.}
Koninin \textbf{y\"uzeyi,}
tepe noktas\i ndan taban\i n s\i n\i r\i na giden do\u grular taraf\i ndan
olu\c sturulur.
Koninin tepe noktas\i ndan taban\i n merkezine giden do\u gru,
koninin \textbf{eksenidir} (\gr{>'axwn} ``dingil'').
Bu eksen, koninin taban\i na dik ise,
koninin kendisi \textbf{diktir.}
Her koni i\c cin,
ekseni i\c ceren her d\"uzlem,
koniyi bir \"u\c cgende keser.
Bu \"u\c cgene \textbf{eksen \"u\c cgeni} denebilir.
\end{definition}

\begin{remark}
Bir koni dik olmayabilir.
Koninin eksen \"u\c cgeninin taban\i,
koninin taban\i n\i n bir \c cap\i d\i r.
\end{remark}

\begin{theorem}\label{thm:cone}
Bir koninin bir eksen \"u\c cgeni,
 \Sekilde{fig:ax-base}ki gibi
\begin{figure}[ht]
\psset{unit=8mm}
\mbox{}\hfill
\begin{pspicture}(0,-0.5)(4,4)
%\psgrid
\pspolygon(0,0)(4,0)(1,4)
\psset{linestyle=dashed}
\psline(1,0)(1.75,3)
\psline(1,0)(2,2.67)
\psline(1,0)(1.5,3.33)
\uput[l](1,4){$A$}
\uput[l](0,0){$B$}
\uput[r](4,0){$C$}
\uput[d](1,0){$F$}
\uput[ur](1.75,3){$G$}
\uput[ur](1.5,3.33){$G$}
\uput[ur](2,2.67){$G$}
\psdots(1,0)
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\hfill\mbox{}
\caption{Koninin eksen \"u\c cgeni ve taban\i}\label{fig:ax-base}
\end{figure}
taban\i\ $BC$ olan $ABC$ \"u\c cgeni olsun.
Koninin taban\i n\i n $DE$ kiri\c si \c cizilsin,
ve bu kiri\c s, $BC$ \c cap\i na dik olsun.
O zaman kiri\c s, 
\c cap taraf\i ndan bir $F$ noktas\i nda ikiye b\"ol\"un\"ur, ve
\begin{equation}\label{eqn:DF^2}
  DF^2=BF\cdot FC.
\end{equation}
\end{theorem}

\begin{definition}\label{def:conic}
\Teorem{thm:cone} durumunda
$DE$ kiri\c sini i\c ceren bir d\"uzlem,
eksen \"u\c cgeninin $AC$ kenar\i n\i\
bir $G$ noktas\i nda kessin.
O zaman bu d\"uzlem,
koninin y\"uzeyini
\Sekilde{fig:curve}ki gibi
\begin{figure}[ht]
\centering
\psset{unit=8mm}
\begin{pspicture}(-2,-0.5)(2,3.5)
%\psgrid
\psTilt{104.04}{
\psplot{-1.732}{1.732}{3 x x mul sub}
\psline(0,0)(0,3)
\psline(-1.414,1)(1.414,1)
}
\psline(-1.732,0)(1.732,0)
\uput[d](-1.732,0){$D$}
\uput[d](1.732,0){$E$}
\uput[d](0,0){$F$}
\uput[u](-0.75,3){$G$}
\uput[l](-1.6,1){$K$}
\uput[r](1.2,1){$L$}
\uput[ur](-0.2,1){$M$}
\end{pspicture}
\caption{Bir koni kesiti}\label{fig:curve}
\end{figure}
bir $DGE$ e\u grisinde keser.
Bu e\u griye \textbf{koni kesiti} denir.
$DE$ do\u grusu, e\u grinin bir kiri\c sidir.
\end{definition}

\begin{theorem}\label{thm:conic}
$KL$ do\u grusu,
yukar\i daki koni kesitinin ba\c ska bir kiri\c si olsun,
ve bu kiri\c s,
$DE$ kiri\c sine paralel olsun.
$KL$ kiri\c si ve $FG$ do\u grusu bir $M$ noktas\i nda kesi\c sir.
Ayr\i ca koninin taban\i na paralel olan
ve $KL$ kiri\c sini i\c ceren bir d\"uzlem vard\i r.
Bu d\"uzlem,
\begin{compactitem}
\item 
$ABC$ \"u\c cgenini
$BC$ taban\i na paralel olan bir $NP$ do\u grusunda keser, ve
\item
koninin kendisini, \c cap\i\ $NP$ olan bir dairede keser.
\end{compactitem}
\Sekle{fig:2bases}
\begin{figure}[ht]
\psset{unit=8mm}
\begin{pspicture}(-0.5,-0.5)(4,4)
%\psgrid
\pspolygon(0,0)(4,0)(1,4)
\psline(0.25,1)(3.25,1)
\psset{linestyle=dashed}
\psline(1,0)(2.00,2.67)
\psline(1,0)(1.75,3.00)
\psline(1,0)(1.50,3.33)
\uput[l](1,4){$A$}
\uput[l](0,0){$B$}
\uput[r](4,0){$C$}
\uput[d](1,0){$F$}
\uput[ur](1.75,3){$G$}
\uput[ur](1.5,3.33){$G$}
\uput[ur](2,2.67){$G$}
\uput[l](0.25,1){$N$}
\uput[r](3.25,1){$P$}
\uput[dr](1.3,1){$M$}
\psdots(1,0)
\end{pspicture}
\hfill
\begin{pspicture}(0,-1)(3.5,3.5)
\pscircle(1.5,2){1.5}
\psline(0,2)(3,2)
\psset{linestyle=dashed}
\psline(0.9,0.636)(0.9,3.364)
\psline(1.0,0.586)(1.0,3.414)
\psline(1.1,0.556)(1.1,3.444)
\uput[l](0,2){$N$}
\uput[r](3,2){$P$}
\uput[dl](1,0.586){$K$}
\uput[ul](1,3.414){$L$}
\uput[ur](1,2){$M$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\caption{Koninin eksen \"u\c cgeni ve tabanlar\i}\label{fig:2bases}
\end{figure}
bak\i n.
Koni kesitinin $LK$ kiri\c si,
bu yeni dairenin kiri\c sidir,
ve dairenin $NP$ \c cap\i na diktir,
dolay\i s\i yla $KM=ML$.
Bu \c sekilde $GF$ \i\c s\i n\i,
$DGE$ koni kesitinin
$DE$ kiri\c sine paralel olan her kiri\c si ikiye b\"oler.
\end{theorem}

\begin{definition}
Tan\i m \ref{def:conic}
ve \Teoremde{thm:conic} 
$G$ noktas\i, koni kesitinin \textbf{k\"o\c sesidir,}
ve $GF$ \i\c s\i n\i\
koni kesitinin bir \textbf{\c cap\i d\i r,}
\c c\"unk\"u $DE$ kiri\c sine paralel olan kiri\c sleri ikiye b\"oler.
E\u ger \c cap,
ikiye b\"old\"u\u g\"u 
ve birbirine paralel olan
kiri\c slere dik ise,
ona \textbf{eksen} denir.
Ama her durumda
$DE$ kiri\c sinin $DF$ (veya $EF$) yar\i s\i na
\textbf{ordinat} denir,
ve \c cap\i n $GF$ par\c cas\i na,
$DF$ ordinat\i na kar\c s\i l\i k gelen \textbf{absis} denir.
\end{definition}

\begin{remark}
O zaman $KM$ ve $LM$ do\u grular\i\ da
ordinatt\i r,
ve onlara kar\c s\i l\i k gelen absis, $GM$ do\u grusudur.
\end{remark}

\begin{theorem}\label{thm:para}
\Sekilde{fig:2bases}ki durumda $FG\parallel BA$ olsun.
O zaman 
\begin{equation*}
GM:GF::ML^2:FE^2.
\end{equation*}
Sonu\c c olarak
bir $\ell$ uzunlu\u gu i\c cin,
koni kesitinin herhangi ordinat\i n\i n uzunlu\u gu $y$
ve ordinata kar\c s\i l\i k gelen absisin uzunlu\u gu $x$ ise
\begin{equation*}
  y^2=\ell x.
\end{equation*}
Ayr\i ca
\begin{equation*}
  \ell:GA::CB^2::CA\cdot CB.
\end{equation*}
\end{theorem}

\begin{definition}
  \Teoremde{thm:para}ki koni kesiti
\textbf{parabold\"ur}
(\gr{parabol'h} ``uygulama, yerle\c stirme''),
ve $\ell$, parabol\"un \textbf{parametresidir}
ve parabol\"un \textbf{dikey kenar\i n\i n} uzunlu\u gudur.%%%%%
\footnote{Dikey kenar\i n Latince'si, \emph{latus rectum.}}
\end{definition}

\begin{definition}
  $a:c::c:d$ ve $c:d::d:b$ ise $c$ ve $d$ uzunluklar\i na
$a$ ve $b$ uzunluklar\i n\i n
\textbf{iki orta orant\i l\i s\i} denir.
\end{definition}

\begin{theorem}[Menaechmus]\nocite{MR2093668}\nocite{MR13:419a}
  Parametreleri $a$ ve $b$ olan paraboller ile
$a$ ve $b$ uzunluklar\i n\i n iki orta orant\i l\i s\i\
bulunabilir.  Asl\i nda
  \begin{equation*}
    a:x::x:y::y:b
  \end{equation*}
orant\i lar\i\
\Sekilde{fig:Men}ki gibi \c c\"oz\"ulebilir.
\begin{figure}[ht]
\psset{unit=2cm}
  \centering
  \begin{pspicture}(0,-0.25)(3,2)
    \psplot02{x x mul 2 div}
\psplot03{x sqrt}
\psline(0,2)(0,0)(3,0)
\psline(0,1.2599)(1.5874,1.2599)(1.5874,0)
\uput[dr](1.5874,1.2599){$A$}
\uput[d](1.5874,0){$B$}
\uput[dl](0,0){$C$}
\uput[l](0,1.2599){$D$}
  \end{pspicture}
  \caption{\.Iki orta orant\i l\i}\label{fig:Men}
  
\end{figure}
Parametresi $b$ olan parabol\"un bir ordinat\i\ $AB$ ve
ona kar\c s\i l\i k gelen absis $CB$ ise, ve
parametresi $a$ olan parabol\"un bir ordinat\i\ $AD$ ve
ona kar\c s\i l\i k gelen absis $CD$ ise,
ve her parabol\"un ordinatlar\i\
di\u ger parabol\"un \c cap\i na paralel ise,
o zaman $CB$ ve $CD$ do\u grular\i n\i n uzunluklar\i\
yukar\i daki orant\i lar\i\ \c c\"ozer.
\end{theorem}

\begin{remark}
  Aristo hakk\i nda yorumlar\i nda Eutocius,
iki orta orant\i l\i\ probleminin birka\c c tane \c c\"oz\"um\"u verdi.
Bunlar\i n biri, yukar\i daki Menaechmus'un \c c\"oz\"um\"uyd\"u.
Aristo'nun ve Eutocius'un metinleri, Miletli \.Isidorus taraf\i ndan topland\i.
\.Isidorus, Ayasofya'n\i n iki mimar\i ndan biriydi.
\end{remark}

\section{Hacimler}

\begin{definition}
  Dik paralely\"uz\"un \textbf{hacmi,}
onun e\c sitlik s\i n\i f\i d\i r.
Geni\c sli\u gi $a$, y\"uksekli\u gi $b$,
ve derinli\u gi $c$ olan dik paralely\"uz\"un hacmi
\begin{equation*}
  a\cdot b\cdot c
\end{equation*}
veya $abc$ ile g\"osterilir.
\end{definition}

\begin{theorem}
  $abc=bac=bca$ ve $ab(c+d)=abc+abd$.
\end{theorem}

\begin{theorem}
  $abc=ade\implies bc=de$.
\end{theorem}

\begin{theorem}
  $ab:cd::e:f\iff abf=cde$.
\end{theorem}

\section{Hiperboller}

\begin{theorem}\label{thm:hyp}
\Sekilde{fig:ax-base} koni kesitinin $GF$ \c cap\i\
$G$ noktas\i n\i n \"otesine uzat\i l\i rsa,
\Sekilde{fig:hyper}ki gibi
\begin{figure}[ht]
\begin{pspicture}(-1,-2)(4,4)
\psline(1,0)(2,4)(0,0)(4,0)(1,2)
\psline(0,0)(-0.5,-1)(5.5,-1)(4,0)
\psline(1,0)(0.744,-1)
\psline[linestyle=dashed](0.744,-1)(2.798,-1.515)(3.476,1.200)(3.076,1.302)
\psline[linestyle=dashed](2.647,-0.412)(2.798,-1.515)
\uput[l](-0.5,-1){$N$}
\uput[d](0.744,-1){$M$}
\uput[ur](5.5,-1){$P$}
\uput[d](2.798,-1.515){$U$}
%\uput[r](3.476,1.200){$V$}
\uput[l](1,2){$A$}
\uput[l](0,0){$B$}
\uput[ur](4,0){$C$}
%\uput{3pt}[120](1,0){$F$}
\uput[ur](1,0){$F$}
\uput[ur](1.438,1.714){$G$}
\uput[l](2,4){$X$}
\psset{linestyle=dashed}
\psline(1,0)(2.647,-0.412)(2,4)
\psline(1.438,1.714)(3.076,1.302)(2.647,-0.412)
\psline(2.370,1.479)(1.941,-0.235)
\psline(2.932,3.765)(3.647,3.588)(3.076,1.302)
\uput[ur](2.370,1.479){$H$}
\uput[dl](2.647,-0.412){$R$}
\uput[ur](3.076,1.302){$S$}
\uput[dl](1.941,-0.235){$T$}
\psline(2.370,1.479)(2.932,3.765)(2,4)
\uput[u](2.932,3.765){$Y$}
\uput[ur](3.647,3.588){$Z$}
\psset{linestyle=dotted}
\psline(0.5,0)(1,2)
\uput[d](0.5,0){$J$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-2)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\psline[linestyle=dashed](1,0.268)(2.732,0.268)(2.732,2)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\caption{Konide hiperbol}\label{fig:hyper}
\end{figure}
$BA$ do\u grusunun uzat\i lmas\i n\i\ bir $X$ noktas\i nda kessin.
$FR$ do\u grusu, $GF$ \c cap\i na dik olsun ve
\begin{equation}\label{eqn:FRFG}
  FR\cdot FG=DF^2
\end{equation}
e\c sitli\u gini sa\u glas\i n.  
$MU\parallel FR$ olsun, ve
(gerekirse uzat\i lm\i\c s)
$XR$ ve $MU$, $U$ noktas\i nda kesi\c ssin.  O zaman
\begin{equation*}
  GM\cdot MU=KM^2.
\end{equation*}
($KM$, \Sekilde{fig:2bases}ki gibidir.)
 $AJ\parallel XF$ olsun;
o zaman
\begin{equation*}
  GH:GX::BJ\cdot JC:AJ^2.
\end{equation*}
$GH$ do\u grusunun uzunlu\u gu $\ell$ olsun,
ve $GX$ do\u grusunun uzunlu\u gu $2a$ olsun.
Koni kesitinin herhangi bir ordinat\i n\i n uzunlu\u gu $y$
ve bu ordinata kar\c s\i l\i k gelen absisin uzunlu\u gu $x$ ise
\begin{equation*}
  2ay^2=2a\ell x+\ell x^2.
\end{equation*}
\end{theorem}

\begin{remark}
\Sekle{fig:hyper-graph}
\begin{figure}[ht]
  \centering
\psset{unit=8mm}
  \begin{pspicture}(-2,-4.4)(6,3.333)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]04{t t t 2 add mul 6 div sqrt add
                                 t t 2 add mul 2 div sqrt}
\parametricplot[linewidth=2pt]05{t t t 2 add mul 6 div sqrt sub
                                 0 t t 2 add mul 2 div sqrt sub}
\psline(1.419,-2.739)(4.581,2.739)
\psline(-2,0)(6,0)
\psdots(0,0)(0,1.333)(-2,0)
\psline(0,0)(0,1.333)
\uput[d](-1,0){$2a$}
\uput[l](0,0.667){$\ell$}
\uput[d](1.5,0){$x$}
\uput[-30](3.791,1.369){$y$}
\uput[-30](2.209,-1.369){$y$}
\psset{linestyle=dotted}
\psline(3,3.333)(-2,3.333)(-2,0)
\psline(-2,1.333)(3,1.333)
\psline(-2,0)(3,3.333)(3,0)
\psline(0,1.333)(0,3.333)
\psline(3,0)(5.739,-1.581)(4.157,-4.320)(1.419,-2.739)
  \end{pspicture}
  \caption{$2ay^2=2a\ell x+\ell x^2$ hiperbol\"u}\label{fig:hyper-graph}
  
\end{figure}
bak\i n; buradaki $\ell$-i\c saretli do\u gru,
koni kesitinin d\"uzlemine dik olarak d\"u\c s\"un\"ulebilir.
\end{remark}

\begin{definition}\sloppy
\Teoremde{thm:hyp}
alan\i\ $y^2$ olan kare,
alan\i\ $\ell x$ olan dikd\"ortgenini a\c st\i\u g\i ndan,
koni kesitine
\textbf{hiperbol} (\gr{<uperbol'h} ``a\c sma'') denir;
$GH$ do\u grusu,
hiperbol\"un \textbf{dikey kenar\i d\i r;}
dikey kenar\i n $\ell$ uzunlu\u gu, hiperbol\"un
\textbf{parametresidir;}
$GX$ do\u grusu, hiperbol\"un \textbf{yanlamas\i na kenar\i d\i r;}%%%%%
\footnote{\gr{pl'agia pleur'a}; Latince'si \emph{latus transversum.}}
%%%%%%%%%%%%%
yanlamas\i na kenar\i n orta noktas\i,
hiperbol\"un \textbf{merkezidir.}
\end{definition}


\begin{remark}
\Sekilde{fig:hyper}
$FS$ ve $FH$ dikd\"ortgenlerinin fark\i\ $TS$ dikd\"ortgendir,
ve bu dikd\"ortgen $GY$ dikd\"ortgenine benzerdir.%%%%%
\footnote{Bu $GY$ dikd\"ortgeni, hiperbol\"un \textbf{\c seklidir}
(\gr{e>~idos}).}
%%%%%%%%%%%%%%%%%
\end{remark}

\section{\.I\c saretli uzunluklar ve elipsler}

\begin{definition}
Bir y\"on ile donat\i lm\i\c s bir do\u gru,
bir \textbf{y\"onl\"u do\u grudur.}
E\u ger $AB$, $A$'dan $B$'ye y\"on ile donat\i l\i rsa,
olu\c san y\"onl\"u do\u gru
\begin{equation*}
  \ydp{AB}
\end{equation*}
bi\c ciminde yaz\i labilir.
$\ydp{AA}$,
\textbf{yoz} veya \textbf{dejenere} y\"onl\"u do\u grudur ve
$A$ noktas\i\ olarak anla\c s\i labilir.
\begin{comment}
  

$A$, $B$, ve $C$ bir do\u gruda ise
\begin{align*}
-\ydp{AB}&=\ydp{BA},&
  \ydp{AB}+\ydp{BC}&=\ydp{AC}.
\end{align*}


\end{comment}
E\u ger $ABDC$ ve $DCEF$ paralelkenar ise,
o zaman
\begin{align*}
  \ydp{AB}&=\ydp{CD},&  \ydp{AB}&=\ydp{EF},
\end{align*}
\"Ozel olarak $\ydp{AA}=\ydp{BB}$.
\end{definition}

\begin{theorem}
Do\u grular\i n paralelli\u gi
ve y\"onl\"u do\u grular\i n e\c sitli\u gi,
denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{definition}\label{def:vec}
  Y\"onl\"u do\u grunun e\c sitlik s\i n\i f\i, \textbf{vekt\"ord\"ur.}
\end{definition}

\begin{definition}
  Her paralellik s\i n\i f\i\ i\c cin bir y\"on \textbf{pozitif,}
di\u ger y\"on \textbf{negatif} olsun.
O zaman her (yoz olmayan) y\"onl\"u do\u gru ya pozitif ya negatiftir.
Bir y\"onl\"u do\u grunun pozitifli\u gi veya negatifli\u gi,
y\"onl\"u do\u grunun \textbf{i\c saretidir.}
\end{definition}

\begin{theorem}
$A$, $B$, ve $C$ bir do\u gruda olsun.
O zaman $\ydp{AB}$ ve $\ydp{BC}$ 
y\"onl\"u do\u grular\i n\i n i\c saretleri ayn\i d\i r 
ancak ve ancak $AB<AC$ ve $BC<AC$.  
\end{theorem}

\begin{theorem}\label{thm:sign}
A\c sa\u g\i daki ko\c sulu sa\u glayan $\mathscr S$ ba\u g\i nt\i s\i\
bir denklik ba\u g\i nt\i s\i d\i r:
$\ydp{AB}\mathrel{\mathscr S}\ydp{CD}$ ancak ve ancak
 $\ydp{AB}$ ve $\ydp{CD}$ 
y\"onl\"u do\u grular\i n\i n i\c saretleri ayn\i\ ve $AB=CD$.
\end{theorem}

\begin{definition}
  \Teoremde{thm:sign}ki denklik ba\u g\i nt\i s\i na g\"ore
bir y\"onl\"u do\u grunun denklik s\i n\i f\i,
\textbf{y\"onl\"u do\u grunun uzunlu\u gudur.}
\end{definition}

\begin{remark}
Bir do\u grunun uzunlu\u gu, yeni tan\i m\i\ alabilir:
$\ydp{AB}$ ve $\ydp{BA}$ y\"onl\"u do\u grular\i n\i n uzunluklar\i n\i n
hangisi pozitif ise, 
$AB$ do\u grusunun uzunlu\u gu olarak al\i nabilir.
Bu tan\i m\i\ ba\c slang\i\c ctan kullanabildik.
\end{remark}

\begin{definition}
K\"u\c c\"uk $a$, $b$, $c$, \lips Latin harfleri,
y\"onl\"u do\u grunun 
uzunlu\u gunu (yani i\c saretli uzunlu\u gunu) g\"osterecek.
Yoz y\"onl\"u do\u grunun uzunlu\u gu,
\begin{equation*}
  0
\end{equation*}
olsun, ve $\ydp{AB}=c$ ise
\begin{equation*}
  -c=\ydp{BA}
\end{equation*}
olsun.
\end{definition}

\begin{theorem}
\.Iki i\c saretli uzunluk toplanabilir, ve tan\i ma g\"ore
$A$, $B$, ve $C$ bir do\u gruda ve
\begin{align*}
  \ydp{AB}&=d,&\ydp{BC}&=e,&\ydp{AC}&=f
\end{align*}
ise
\begin{equation*}
  d+e=f.
\end{equation*}
Bu durumda toplama de\u gi\c smeli ve birle\c smelidir;
ayr\i ca
\begin{gather*}
a+0=a,\\
a+(-a)=0.
\end{gather*}
\end{theorem}

\begin{definition}
  $\begin{gathered}[t]
  a-b=a+(-b),\\
-a\cdot b=-(ab)=a\cdot(-b),\\
-a\cdot bc=a\cdot(-b)\cdot c=ab\cdot(-c)=-(abc).
  \end{gathered}$
\end{definition}

\c Simdi hiperbol\"un $2ay^2=2a\ell x+\ell x^2$ denkleminde 
$x$ ve $y$ negatif olabilir.
Ayr\i ca $a$ negatif olabilir, 
ama bu durumda tan\i mlanan e\u gri hiperbol de\u gildir:

\begin{definition}
$\ell>0$ ve $a>0$ ise
  \begin{equation*}
2ay^2=2a\ell x-\ell x^2
  \end{equation*}
  denklemi,
\textbf{dikey kenar\i n\i n} uzunlu\u gu $\ell$ olan,
\textbf{yanlamas\i na kenar\i n\i n} uzunlu\u gu $2a$ olan 
\textbf{elipsi} (\gr{>'elleiyis} ``eksiklik'') tan\i mlar
(ama ordinatlar\i n \c capa a\c c\i s\i n\i\ se\c cilmeli).
\Sekle{fig:ell-graph}
\begin{figure}[ht]
  \centering
\psset{unit=3cm}
  \begin{pspicture}(0,-1.051)(2,1.5)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]02{t t 2 t sub mul 6 div sqrt add
                                   t 2 t sub mul 2 div sqrt}
\parametricplot[linewidth=2pt]02{t t 2 t sub mul 6 div sqrt sub
                                 0 t 2 t sub mul 2 div sqrt sub}
\psline(0.9484,-0.6667)(1.7182,0.6667)
\psline(0,0)(2,0)
\psdots(0,0)(0,1.333)(2,0)%(1,0)
\psline(0,0)(0,1.333)
\pcline[offset=8pt](0,1.3333)(2,1.3333)
\ncput*{$2a$}
%\uput[d](1,0){$2a$}
\pcline[offset=8pt](0,0)(0,1.333)
\ncput*{$\ell$}
%\uput[l](0,0.667){$\ell$}
\uput[d](0.6667,0){$x$}
\uput[150](1.141,-0.333){$y$}
\psset{linestyle=dotted}
\psline(0,1.333)(2,0)
\psline(1.3333,0)(1.3333,1.3333)
\psline(2,0.4444)(0,0.4444)
\psline(0,1.3333)(2,1.3333)(2,0)
\psline(1.3333,0)(2,-0.385)(1.615,-1.051)(0.9484,-0.6667)
  \end{pspicture}
  \caption{$2ay^2=2a\ell x-\ell x^2$ elipsi}\label{fig:ell-graph}
  
\end{figure}
bak\i n.
Hiperboldeki gibi elipsin \textbf{merkezi,}
yanlamas\i na kenar\i n\i n orta noktas\i d\i r.
Hiperbol ve elips, \textbf{merkezli koni kesitidir.}
\end{definition}

\begin{theorem}
\Teoremde{thm:hyp}
%\Sekilde{fig:ax-base} 
koni kesitinin $GF$ \c cap\i\
$F$ noktas\i n\i n \"otesine uzat\i l\i rsa
ve $AB$ do\u grusunun uzat\i lmas\i n\i\ keserse,
hiperbol\"un yerine
elips \c c\i kar.
\end{theorem}

\begin{remark}
\c Simdi her koni kesiti ya parabol ya hiperbol ya da elipstir.
Pergeli Apollonius bu adlar\i\ vermi\c stir.
Parabol olmayan her koni kesiti merkezlidir.
\end{remark}

\chapter{Eksenler}

\section{Eksenler}

\begin{definition}\label{def:axes}
D\"uzlemde iki do\u gru bir $O$ noktas\i nda kesi\c ssin.
Do\u grular\i n birine \textbf{$x$ ekseni,}
di\u gerine \textbf{$y$ ekseni} densin,
ve $O$ noktas\i na \textbf{ba\c slang\i\c c noktas\i} densin.
\end{definition}

\begin{theorem}\label{thm:coord}
$Xy$ eksenleriyle donat\i lm\i\c s d\"uzlemde
her $A$ noktas\i\ i\c cin
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
$ABOC$ paralelkenard\i r.%%%%%
\footnote{E\u ger $A$ zaten bir eksendeyse
$ABOC$ paralelkenar\i\ ``dejenere'' olacakt\i r.
\"Orne\u gin
$A$, $x$ eksenindeyse $B$, $A$ noktas\i d\i r ve
$C$, $O$ noktas\i d\i r.}
%%%%%%%%%%%%%%%%%%%%
Tam tersine $b$ ve $c$ i\c saretli uzunluk olmak \"uzere, 
herhangi bir $(b,c)$ s\i ral\i\ ikilisi i\c cin,
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
d\"uzlemde bir ve tek bir $A$ i\c cin
\begin{align*}
  \ydp{OB}&=b,&\ydp{OC}&=c,
\end{align*}
ve $ABOC$ paralelkenard\i r.
\end{theorem}

\begin{definition}\label{def:coord}
  \Teoremde{thm:coord} $b$, $A$ noktas\i n\i n \textbf{$x$ koordinat\i d\i r,}
ve $c$, $A$ noktas\i n\i n \textbf{$y$ koordinat\i d\i r.}
\Sekilde{fig:axes}ki gibi
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-4,-0.866)(1,2.598)
    \psline{->}(-4,0)(1.5,0)
\psline{->}(-0.5,-0.866)(1.5,2.598)
\psline[linestyle=dotted](1,1.732)(-2,1.732)(-3,0)
\uput[120](-2,1.732){$A$}
\uput[120](1,1.732){$C$}
\uput[-30](1,1.732){$c$}
\uput[120](-3,0){$B$}
\uput[d](-3,0){$b$}
\uput[-60](0,0){$O$}
\uput[ur](1.5,0){$x$}
\uput[r](1.5,2.598){$y$}
  \end{pspicture}
  \caption{Koordinatlar}\label{fig:axes}
  
\end{figure}
$B$ noktas\i na $b$ yaz\i labilir, ve
$C$ noktas\i na $c$ yaz\i labilir.
\end{definition}

\begin{remark}
\Sekilde{fig:2branches}ki
\begin{figure}[ht]
  \centering
\psset{unit=7.5mm}
  \begin{pspicture}(-8,-4)(6,4)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]04{t t t 2 add mul 6 div sqrt add
                                   t t 2 add mul 2 div sqrt}
\parametricplot[linewidth=2pt]05{t t t 2 add mul 6 div sqrt sub
                                 0 t t 2 add mul 2 div sqrt sub}
\parametricplot[linewidth=2pt]04{-2 t t t 2 add mul 6 div sqrt add sub
                                  0 t t 2 add mul 2 div sqrt sub}
\parametricplot[linewidth=2pt]05{-2 t t t 2 add mul 6 div sqrt sub sub
                                 t t 2 add mul 2 div sqrt}
\psline{->}(-8,0)(6,0)
\uput[d](6,0){$x$}
\psline{->}(-2.309,-4)(2.309,4)
\uput[ul](2.309,4){$y$}
\psdots(0,0)(-2,0)%(0,1.333)
%\psline(0,0)(0,1.333)  % latus rectum
\uput[dr](-2,0){$-2a$}
\uput[120](0,0){$O$}
%\uput[l](0,0.667){$\ell$}
\psdots(3,0)(-5,0)(1.581,2.739)(-1.581,-2.739)
\uput[dr](3,0){$b$}
\uput[dr](-5,0){$-2a-b$}
\uput[120](1.581,2.739){$c$}
\uput[dr](-1.581,-2.739){$-c$}
\psset{linestyle=dotted}
\pspolygon(1.419,-2.739)(4.581,2.739)(-3.419,2.739)(-6.581,-2.739)
  \end{pspicture}
  \caption{\.Ikinci dal\i\ ile $2ay^2=2a\ell x+\ell x^2$ hiperbol\"u}\label{fig:2branches}
  
\end{figure}
koordinatlar\i\ $(b,c)$ olan nokta hiperboldeyse,
koordinatlar\i\ $(b,-c)$, $(-2a-b,c)$, ve $(-2a-b,-c)$ olan noktalar\i\ da
hiperboldedir.
\end{remark}

\begin{theorem}
  Denklemi $2ay^2=2a\ell x+\ell x^2$ olan hiperbol\"u verilsin,
ama yeni $st$ eksenleri se\c cilsin.
E\u ger
\begin{compactitem}
\item
  $s$ ekseni, $x$ eksenidir, ve
\item
$t$ ekseni, hiperbol\"un merkezinden ge\c cer ve $y$ eksenine paralel ise,
\end{compactitem}
o zaman yeni $st$ eksenlerine g\"ore hiperbol\"un denklemi,
\begin{equation*}
  2at^2=\ell s^2-\ell a^2.
\end{equation*}
\end{theorem}

\begin{theorem}\label{thm:hyp-cen}
\.I\c saretli uzunluklar\i n oran\i
  \begin{equation*}
    a:b::c:d\iff ad=bc
  \end{equation*}
kural\i na g\"ore tan\i mlanabilir.
Oranlar\i n toplam\i
\begin{equation*}
  (a:c)+(b:c)::(a+b):c
\end{equation*}
kural\i na g\"ore tan\i mlanabilir.
\end{theorem}

\begin{remark}
  \c Simdi oranlar\i\ say\i lar gibi kullanabiliriz.
\end{remark}

\begin{definition}
  $a:a$ oran\i
  \begin{equation*}
    1
  \end{equation*}
olarak yaz\i ls\i n, ve $a:b$ oran\i
\begin{equation*}
  \frac ab
\end{equation*}
veya $a/b$ bi\c ciminde yaz\i ls\i n.
O zaman hiperbol\"un
\begin{equation*}
2ay^2=2a\ell x+\ell x^2
\end{equation*}
denklemi
\begin{equation*}
  y^2=\ell x+\frac{\ell}{2a}x^2
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu denkleme \textbf{Apollonius denklemi} diyelim.
\Teoreme{thm:hyp-cen} g\"ore, 
farkl\i\ eksenlere g\"ore,
hiperbol\"un
\begin{equation*}
2ay^2=\ell x^2-\ell a^2
\end{equation*}
denklemi de vard\i r;
bu denklem
\begin{equation*}
  \frac{x^2}{a^2}-\frac{y^2}{\ell a/2}=1
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu denkleme \textbf{merkez denklemi} diyelim.
\end{definition}

\section{Dik eksenler}


\begin{theorem}
  Parabolde \c capa paralel olan her do\u gru, yeni bir \c capt\i r.
\Sekilde{fig:parab-new}ki gibi
\begin{figure}[ht]

\centering
\psset{unit=4cm}
\begin{pspicture}(-0.5625,-0.1)(1.5625,1.35)
\psline(-0.5625,0)(1.5625,0) % old x axis
%\psline(0,0)(-1.25,1.25) % old y axis
\psline(0.8125,0.75)(-0.1875,0.75)%(-0.75,0.75) % new x axis
\psline(-0.5625,0)(-0.1875,0.75)%(0.1825,1.5) % new y axis
\parametricplot{0}{1.25}{t t mul t sub
                       t} % parabola x = y^2 - y 
\psline(-0.1825,0.75)(0.5625,0) % old ordinate of new origin
\psline(0.0625,0.75)(0.3125,1.25) % new ordinate of point
\psline(0.3125,1.25)(1.5625,0) % old ordinate of point
\uput[d](0,0){$A$}
\uput[ul](-0.1875,0.75){$B$}
\uput[u](0.3125,1.25){$C$}
\uput[d](0.5625,0){$D$}
\uput[d](1.5625,0){$E$}
\uput[d](-0.5625,0){$F$}
\uput[d](0.0625,0.75){$H$}
\uput[ur](0.8125,0.75){$G$}
\uput[u](-0.28125,0){$a$}
\uput[u](0.28125,0){$a$}
\uput[u](1.0625,0){$x-a$}
\uput[ul](-0.375,0.375){$c$}
\uput[ur](0.1875,0.375){$b$}
\uput[ur](1.1875,0.375){$b$}
\uput[u](-0.0625,0.75){$s$}
\uput[u](0.4375,0.75){$x-a-s$}
\uput[dr](0.1875,1){$t$}
\uput[ur](0.5625,1){$y-b$}
  \end{pspicture}
  
  \caption{Parabol\"un yeni \c cap\i}\label{fig:parab-new}
  
\end{figure}
\begin{compactenum}[1)]
  \item
$ABC$ e\u grisi, \c cap\i\ $FE$ ve k\"o\c sesi $A$ olan parabol,
\item
$BD$ ve $CE$ ordinat,
\item
$FA=AD$, ve 
\item
$BG\parallel FE$,
$CH\parallel BF$ 
\end{compactenum}
olsun.
A\c sa\u g\i daki i\c saretli uzunluklar\i\ tan\i mlans\i n:
\begin{align*}
  &
  \begin{gathered}
    \ydp{AD}=a,\\
\ydp{DB}=b,
  \end{gathered}&
&\begin{gathered}
 \ydp{AE}=x,\\
\ydp{EC}=y,   
  \end{gathered}&
&\begin{gathered}
 \ydp{BH}=s,\\
\ydp{HC}=t,   
  \end{gathered}&
\ydp{FB}&=c.
\end{align*}
Parabol\"un dikey kenar\i n\i n uzunlu\u gu $\ell$ ise
\begin{equation*}
  \frac m{\ell}=\frac{c^2}{b^2}
\end{equation*}
olsun.  O zaman
\begin{equation*}
  y^2=\ell x
\end{equation*}
oldu\u gundan
\begin{equation*}
  t^2=ms.
\end{equation*}
\end{theorem}

\begin{theorem}
  Parabol\"un bir (ve tek bir) \c cap\i\ i\c cin
ordinatlar \c capa diktir
(yani Tan\i m \numarada{def:axis}ki gibi parabol\"un ekseni
ve tek bir ekseni vard\i r).
\end{theorem}

\begin{theorem}
  Hiperbol\"un merkezinden ge\c cen ve hiperbol\"u kesen her do\u gru, 
hiperbol\"un yeni bir \c cap\i d\i r.
  \Sekilde{fig:hyper-new}ki gibi
  \begin{figure}[ht]
    \centering
    \psset{unit=3cm}
    \begin{pspicture}(0,-0.167)(2.6,2.567)
\psline(0,0)(2.6,0)     % old diameter, slope 0
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex, slope -2
\psline(1.4,2.4)(2.6,0)     % old ordinate of point
\psline(1.4,0.933)(1.866,0)
\psline(1.4,0.933)(2.133,0.933)
\parametricplot{0}{2.4}{t t mul 1 add sqrt t 2 div sub t}
\psline(0,0)(1.4,0.933)     % new diameter, slope 2/3
\psline(1.4,2.4)(1.4,0.933) % new ordinate of point
\psline(0.866,0.577)(0.866,0)
\uput[d](1,0){$A$}
\uput[ul](0.866,0.577){$B$}
\uput[u](1.4,2.4){$C$}
\uput[d](0,0){$D$}
\uput[d](1.155,0){$E$}
\uput[d](2.6,0){$F$}
\uput[d](0.866,0){$G$}
\uput[ul](1.4,0.933){$H$}
\uput[ur](2.133,0.933){$K$}
\uput[d](1.866,0){$L$}
    \end{pspicture}
    \caption{Hiperbol\"un yeni \c cap\i}\label{fig:hyper-new}
    
  \end{figure}
  \begin{compactenum}[1)]
    \item
  $ABC$ e\u grisi, merkezi $D$ olan ve \c cap\i\ $DF$ olan hiperbol,
\item
$BE$ ve $CF$ ordinat,
\item
$DG:DA::DA:DE$,
\item
$CH\parallel BG$, $HK\parallel DA$, $HL\parallel BE$
  \end{compactenum}
olsun.
A\c sa\u g\i daki i\c saretli uzunluklar tan\i mlans\i n:
\begin{align*}
&\begin{gathered}
\ydp{DF}=x,\\
\ydp{FC}=y,
 \end{gathered}&
&\begin{gathered}
\ydp{DH}=s,\\
\ydp{HC}=t,
 \end{gathered}&
&\begin{gathered}
\ydp{DE}=c,\\
\ydp{EB}=d,
 \end{gathered}&
&\begin{gathered}
\ydp{DA}=a,\\
\ydp{GE}=e,
 \end{gathered}&
&\begin{gathered}
\ydp{DB}=f,\\
\ydp{GB}=g.
 \end{gathered}
\end{align*}
(\Sekle{fig:hyper-detail} bak\i n.)
\begin{figure}[ht]
    \psset{unit=3cm,labelsep=2pt}
%\mbox{}\hfill
    \begin{pspicture}(0,-0.167)(1.866,2.567)
\psline(0,0)(1.866,0)     % old diameter
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex
\psline(1.4,0.933)(1.866,0)
\psline(0,0)(1.4,0.933)     % new diameter
\uput[207](1.01,0.288){$d$}
\uput[-56](0.433,0.288){$f$}
\pcline[offset=8pt](0,0)(1.4,0.934)
\ncput*{$s$}
%\uput[124](0.7,0.467){$s$}
\uput[27](1.633,0.467){$%\displaystyle
\frac dfs$}
\uput[u](0.577,0){$c$}
\pcline[offset=10pt](1.867,0)(0,0)
\ncput*{$%\displaystyle
\frac cfs$}
%\uput[d](0.933,0){$\displaystyle\frac cfs$}
    \end{pspicture}
\hfill
    \begin{pspicture}(0.866,-0.167)(2.133,2.567)
\psline(0.866,0)(1.866,0)     % old diameter
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex
\psline(1.4,2.4)(2.133,0.933)     % old ordinate of point
\psline(1.4,0.933)(1.866,0)
\psline(1.4,0.933)(2.133,0.933)
\psline(0.866,0.577)(1.4,0.933)     % new diameter
\psline(1.4,2.4)(1.4,0.933) % new ordinate of point
\psline(0.866,0.577)(0.866,0)
\uput[27](1.01,0.288){$d$}
\uput[l](0.866,0.288){$g$}
\uput[l](1.4,1.667){$t$}
\uput[d](0.990,0){$e$}
\uput[d](1.767,0.933){$%\displaystyle
\frac egt$}
\uput[27](1.767,1.667){$%\displaystyle
\frac dgt$}
    \end{pspicture}
%\hfill\mbox{}
    \caption{Hiperbol\"un benzer \"u\c cgenleri}\label{fig:hyper-detail}
    
  \end{figure}
Hiperbol\"un dikey kenar\i n\i n uzunlu\u gu $\ell$
ve $2b^2=\ell a$ ise
\begin{equation*}
  \frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\end{equation*}
oldu\u gundan
\begin{equation*}
\frac{s^2}{f^2}-\frac{t^2}{g^2c/e}=1.
\end{equation*}
\end{theorem}

\begin{theorem}
  Hiperbol\"un bir (ve tek bir) \c cap\i\ i\c cin
ordinatlar \c capa diktir,
yani hiperbol\"un bir (ve tek bir) ekseni vard\i r.
\end{theorem}

\section{Uzakl\i k}

\begin{remark}
  Dik \"u\c cgenle $x^2=a^2+b^2$ denkleminin \c c\"oz\"um\"u 
bulunabilir.
\end{remark}

\begin{definition}
  $x^2=a^2+b^2$ denkleminin (pozitif) \c c\"oz\"um\"u
  \begin{equation*}
    \sqrt{a^2+b^2}.
  \end{equation*}
\end{definition}

\begin{definition}
  Eksenler verilirse,
``koordinatlar\i\ $(a,b)$ olan nokta'' ifadesinin yerine
``$(a,b)$ noktas\i'' diyebiliriz.
\end{definition}

\begin{theorem}
  Eksenler dik ise $(a,b)$ noktas\i n\i n $(c,d)$ noktas\i ndan uzakl\i\u g\i
  \begin{equation*}
    \sqrt{(a-c)^2+(b-d)^2}.
  \end{equation*}
\end{theorem}

\begin{definition}\label{def:slope}
  Eksenler dik ve $a\neq c$ ise ucu $(a,b)$ ve $(c,d)$ olan do\u grunun
\textbf{e\u gimi}
\begin{equation*}
  \frac{b-d}{a-c}.
\end{equation*}
\end{definition}

\begin{remark}
Tan\i m \numarada{def:slope} 
eksenlerin dik olmas\i\ gerekmez ama normaldir.
\end{remark}

\begin{theorem}
  Paralel do\u grular\i n e\u gimleri ayn\i d\i r.
Dik eksene g\"ore,
$a\neq c$ ise $(a,b)$ ve $(c,d)$ 
noktalar\i ndan ge\c cen u\c csuz do\u grunun noktalar\i,
\begin{equation*}
  y=\frac{d-b}{c-a}\cdot(x-a)+b
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
E\u gimi $e/f$ olan ve $(a,b)$ noktas\i ndan ge\c cen u\c csuz do\u grunun
noktalar\i,
\begin{equation*}
  y=\frac ef\cdot(x-a)+b
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
$y$ eksenine paralel olan ve $(a,b)$ noktas\i ndan ge\c cen u\c csuz do\u grunun
noktalar\i,
\begin{equation*}
  x=a
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
\end{theorem}

\begin{remark}
\c Su anda Descartes'\i n ortaya koydu\u gu uyla\c s\i m uygundur:
\end{remark}

\begin{definition}
Bir \textbf{birim} uzunlu\u gu se\c cilirse,
\begin{equation*}
  1
\end{equation*}
olarak yaz\i labilir.
E\u ger $a\cdot b=c\cdot 1$ ise,
o zaman $ab$ alan\i\ $c$ olarak anla\c s\i labilir.
Bu \c sekilde alan, hacim, oran---her \c sey bir uzunluk olur.
\"Ozel olarak e\u gim, bir harf ile yaz\i labilir.
\end{definition}

\begin{theorem}
  Dik eksenlere ve birim uzunlu\u guna g\"ore
$y$ eksenine paralel olmayan do\u grunun denkli\u gi
\begin{equation*}
  y=mx+b
\end{equation*}
bi\c ciminde yaz\i labilir, ve bunun gibi her denklem,
e\u gimi $m$ olan ve $(0,b)$ noktas\i ndan ge\c cen do\u gruyu tan\i mlar.
Benzer \c sekilde $a\neq0$ veya $b\neq0$ ise
(yani $a^2+b^2\neq0$ ise)
\begin{equation*}
  ax+by+c=0
\end{equation*}
denklemi bir do\u gru tan\i mlar,
ve her do\u grunun denklemi bu \c sekilde yaz\i labilir.
\end{theorem}

\begin{definition}
  \Sekilde{fig:cos} $\angle BAC$ dik ise
%$ABC$ \"u\c cgeninde $B$ a\c c\i s\i\ dik ve
%  \begin{align*}
%    AB&=c,&AC&=b,&\angle BAC&=\alpha
%  \end{align*}
%ise
\begin{equation*}
  \cos\alpha=\frac bc.
\end{equation*}%
\begin{figure}[ht]
\psset{unit=8mm}
\centering
\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(2,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](2,2){$C$}
\uput[r](2,1){$a$}
\uput[ul](1,1){$b$}
\uput[22.5](0,0){$\alpha$}
\end{pspicture}
    \caption{Kosin\"us tan\i m\i}\label{fig:cos}
    
  \end{figure}
Burada $\alpha$, $\angle BAC$ a\c c\i s\i n\i n e\c sitlik s\i n\i f\i\ olarak
anla\c s\i labilir,
ve $\cos\alpha$, a\c c\i n\i n \textbf{kosin\"us\"ud\"ur.}
Dik a\c c\i n\i n \"ol\c c\"us\"u
\begin{equation*}
  \frac{\uppi}2.
\end{equation*}
O zaman
\begin{equation*}
  \cos\frac{\uppi}2=0,
\end{equation*}
ve $\beta$ geni\c s a\c c\i\ ise
\begin{equation*}
  \cos\beta=-\cos(\uppi-\beta).
\end{equation*}
\end{definition}

\begin{theorem}[Kosin\"us Teoremi]\label{thm:cos}
  \Sekilde{fig:tri}
  \begin{equation*}
    a^2=b^2+c^2-2bc\cos\alpha.
  \end{equation*}
  \begin{figure}[ht]
\psset{unit=8mm}
\mbox{}\hfill
\begin{pspicture}(-1,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(-1,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](-1,2){$C$}
\uput[ur](0.5,1){$a$}
\uput[dl](-0.5,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(0,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](0,2){$C$}
\uput[ur](1,1){$a$}
\uput[l](0,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill
\begin{comment}
  

\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(1,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](1,2){$C$}
\uput[ur](1.5,1){$a$}
\uput[ul](0.5,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill


\end{comment}
\begin{pspicture}(0,-0.5)(3,2.5)
  \pspolygon(0,0)(2,0)(3,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](3,2){$C$}
\uput[dr](2.5,1){$a$}
\uput[ul](1.5,1){$b$}
\uput{9pt}[18](0,0){$\alpha$}
\end{pspicture}
\hfill\mbox{}
    \caption{Kosin\"us Teoremi}\label{fig:tri}
    
  \end{figure}
\end{theorem}

\begin{definition}
  Dik eksenlere g\"ore
  \begin{gather*}
    (a,b)\cdot(c,d)=ac+bd,\\
\lVert(a,b)\rVert=\sqrt{(a,b)\cdot(a,b)}=\sqrt{a^2+b^2}.
  \end{gather*}
$(a,b)$ noktas\i, $(0,0)$ ba\c slang\i\c c noktas\i ndan
$(a,b)$ noktas\i na giden y\"onl\"u do\u gru olarak anla\c s\i labilir.
\end{definition}

\begin{theorem}
  Dik eksenlere g\"ore $(a,b)$ ve $(c,d)$ aras\i ndaki a\c c\i\ $\theta$ ise
  \begin{equation*}
    (a,b)\cdot(c,d)=\lVert(a,b)\rVert\cdot\lVert(c,d)\rVert\cdot\cos\theta.
  \end{equation*}
\end{theorem}

\begin{theorem}[Cauchy--Schwartz E\c sitsizli\u gi]
  \begin{equation*}
(ac+bd)^2\leq(a^2+b^2)\cdot(c^2+d^2).
  \end{equation*}
\end{theorem}

\begin{definition}[mutlak de\u ger]
  $\abs a=
  \begin{cases}
    a,&\text{ e\u ger $a\geq0$ ise},\\
-a,&\text{ e\u ger $a<0$ ise}.
  \end{cases}$
\end{definition}

\begin{theorem}
  Dik eksenlere ve birim uzunlu\u guna g\"ore
$(s,t)$ noktas\i n\i n $ax+by+c=0$ do\u grusuna uzakl\i\u g\i
  \begin{equation*}
    \frac{\abs{as+bt+c}}{\sqrt{a^2+b^2}}.
  \end{equation*}
\end{theorem}

\section{Dik eksenlere g\"ore koni kesitleri}

\begin{definition}
  $a\neq0$ ise
  \begin{equation*}
    \frac a{\infty}=0.
  \end{equation*}
\end{definition}

\begin{remark}
\c Simdi $\ell>0$ ve $a\neq0$ ise,
dik eksenlere g\"ore,
\begin{equation*}
  y^2=\ell x-\frac{\ell}{2a}x^2
\end{equation*}
Apollonius denklemi,
ekseni $x$ ekseni olan
ve k\"o\c sesi ba\c slang\i\c c noktas\i\ olan
\begin{itemize}
\item 
$a<0$ durumunda hiperbol\"u,
\item
$a=\infty$ durumunda parabol\"u,
\item
$a>0$ durumunda elipsi
\end{itemize}
tan\i mlar.
\Sekle{fig:all} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=1.5cm,plotpoints=200,labelsep=3pt}
  \begin{pspicture}(-3,-2.74)(3,2.74)
\psline{->}(-3,0)(3,0)
\psline{->}(0,-2.74)(0,2.74)
\psset{linewidth=1.6pt}
    \psplot03{x x x mul 2 div add sqrt}
    \psplot03{x x x mul 2 div add sqrt neg}
    \psplot{-3}{-2}{x x x mul 2 div add sqrt}
    \psplot{-3}{-2}{x x x mul 2 div add sqrt neg}
    \psplot0{2.3}{x x x mul 1 div add sqrt}
    \psplot0{2.3}{x x x mul 1 div add sqrt neg}
    \psplot{-3}{-1}{x x x mul 1 div add sqrt}
    \psplot{-3}{-1}{x x x mul 1 div add sqrt neg}
    \psplot03{x                   sqrt}
    \psplot03{x                   sqrt neg}
    \psplot02{x x x mul 2 div sub sqrt}
    \psplot02{x x x mul 2 div sub sqrt neg}
    \psplot01{x x x mul 1 div sub sqrt}
    \psplot01{x x x mul 1 div sub sqrt neg}
\uput[dr](1,0){$a=1/2$}
\uput[ur](2,0){$a=1$}%(1.75,0.4677)
\uput[dr](2,1.414){$a=\infty$}
\uput[r](2,2){$a=-1$}
\uput[ul](2,2.4495){$a=-1/2$}
  \end{pspicture}
  \caption{$y^2=x-x^2/2a$ koni kesitleri}\label{fig:all}
  
\end{figure}
\end{remark}

\begin{remark}
  $\ell>0$ ve $a\neq0$ (ve $a\neq\infty$) olsun, ve
  \begin{align*}
    b&>0,&2b^2=\ell a
  \end{align*}
olsun.
O zaman dikey kenar\i\ $\ell$ olan,
yanlamas\i na kenar\i\ $2\abs a$ olan merkezli koni kesitlerinin 
merkez denklemi
\begin{equation*}
  \frac{x^2}{a^2}\pm\frac{y^2}{b^2}=1.
\end{equation*}
\Sekle{fig:central} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=1.2cm,plotpoints=200}
  \begin{pspicture}(-4,-2)(4,2)
\psline{->}(-4,0)(4,0)
\psline{->}(0,-2)(0,2)
\psset{linestyle=dotted}
\psline(-4,-2)(4,2)
\psline(-4,2)(4,-2)
\pspolygon(-2,-1)(-2,1)(2,1)(2,-1)
\psset{linestyle=solid,linewidth=1.6pt}
    \psplot{-2}2{1 x x mul 4 div sub sqrt}
    \psplot{-2}2{1 x x mul 4 div sub sqrt neg}
    \psplot24{1 x x mul 4 div sub neg sqrt}
    \psplot24{1 x x mul 4 div sub neg sqrt neg}
    \psplot{-2}{-4}{1 x x mul 4 div sub neg sqrt}
    \psplot{-2}{-4}{1 x x mul 4 div sub neg sqrt neg}
\uput[ur](2,0){$a$}
\uput[dl](-2,0){$-a$}
\uput[ur](0,1){$b$}
\uput[ur](0,-1){$-b$}
  \end{pspicture}
  \caption{$x^2/4\pm y^2=1$ koni kesitleri}\label{fig:central}
  
\end{figure}
\end{remark}

\begin{definition}
$x^2/a^2-y^2/b^2=0$ denklemi,  
$x^2/a^2-y^2/b^2=1$ hiperbol\"un \textbf{asimptotlar\i n\i} tan\i mlar.
Yani hiperbol\"un asimptotlar\i, $y=\pm(b/a)x$ do\u grular\i d\i r.
\end{definition}

\begin{theorem}
  $y^2=\ell x+(\ell/2a)x^2$ hiperbol\"un asimptotlar\i n\i n denklemi
  \begin{equation*}
    y=\pm\sqrt{\frac{\ell}{2a}}\cdot(x-a).
  \end{equation*}
\end{theorem}

\begin{definition}
Denklemi $y^2=\ell x$ olan parabol\"un \textbf{odak noktas\i}
\begin{equation*}
  \left(\frac{\ell}4,0\right)
\end{equation*}
ve \textbf{do\u grultman do\u grusu}
\begin{equation*}
  x+\frac{\ell}4=0.
\end{equation*}
\end{definition}

\begin{theorem}
  Denklemi $y^2=\ell x$ olan parabol\"un noktalar\i,
odak noktas\i na ve do\u grultmana uzakl\i\u g\i\ ayn\i\ olan noktalard\i r.
\end{theorem}

\begin{definition}
Denklemi $x^2/a^2-y^2/b^2=1$ olan hiperbol\"un
\textbf{odak noktalar\i}
\begin{equation*}
  (\pm\sqrt{a^2+b^2},0),
\end{equation*}
(s\i ras\i yla) \textbf{do\u grultman do\u grular\i}
\begin{equation*}
  x=\pm\frac{a^2}{\sqrt{a^2+b^2}},
\end{equation*}
ve \textbf{d\i\c smerkezlili\u gi}
\begin{equation*}
  \frac{\sqrt{a^2+b^2}}a.
\end{equation*}
\Sekle{fig:hyp-foc} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=5mm,plotpoints=200}
  \begin{pspicture}(-9,-6)(9,6)
%\psset{linewidth=1pt}
\psline{->}(-9,0)(9,0)
\psline{->}(0,-6)(0,6)
\psset{linestyle=dotted}
%\psline(-4.5,-6)(4.5,6) % asymptote
%\psline(-4.5,6)(4.5,-6)
\psline(3,0)(0,5)
\pscircle(0,0)3
\psline(0,3)(1.8,0)
\pspolygon(-3,-4)(-3,4)(3,4)(3,-4)
\psdots(5,0)(-5,0)(3,0)(-3,0)(0,4)(0,-4)
\pscircle(0,0)5
\psset{linestyle=solid,linewidth=1.6pt}
\psline(1.8,-6)(1.8,6) % directrix
\psline(-1.8,-6)(-1.8,6)
    \psplot3{5.4}{x x mul 9 div 1 sub sqrt 4 mul}
    \psplot3{5.4}{x x mul 9 div 1 sub sqrt 4 mul neg}
    \psplot{-5.4}{-3}{x x mul 9 div 1 sub sqrt 4 mul}
    \psplot{-5.4}{-3}{x x mul 9 div 1 sub sqrt 4 mul neg}
\uput[ur](3,0){$a$}
\uput[ul](-3,0){$-a$}
\uput[dl](0,4){$b$}
\uput[ul](0,-4){$-b$}
\uput[dr](5,0){$\sqrt{a^2+b^2}$}
\uput[dl](-5,0){$-\sqrt{a^2+b^2}$}
  \end{pspicture}  
  \caption{Hiperbol\"un odaklar\i\ ve do\u grultmanlar\i}\label{fig:hyp-foc}
  
\end{figure}
Ayr\i ca $0<b<a$ ise
denklemi $x^2/a^2+y^2/b^2=1$ olan elipsin
\textbf{odak noktalar\i}
\begin{equation*}
  (\pm\sqrt{a^2-b^2},0),
\end{equation*}
(s\i ras\i yla) \textbf{do\u grultman do\u grular\i}
\begin{equation*}
  x=\pm\frac{a^2}{\sqrt{a^2-b^2}},
\end{equation*}
ve \textbf{d\i\c smerkezlili\u gi}
\begin{equation*}
  \frac{\sqrt{a^2-b^2}}a.
\end{equation*}
\Sekle{fig:ell-foc} bak\i n.
\begin{figure}[ht]
\psset{unit=1.15cm,plotpoints=200}
  \centering
  \begin{pspicture}(-4.5,-2)(4.5,2)
\psline{->}(-4.5,0)(4.5,0)
\psline{->}(0,-2)(0,2)
\psset{linestyle=dotted}
\pspolygon(-2,-1.732)(2,-1.732)(2,1.732)(-2,1.732)
\psarc(0,1.732)2{180}{360}
\psline(1,0)(1,1.732)
\psarc(0,1.732)1{180}{360}
\psline(0,0.732)(2,1.732)
\psline(0,-0.268)(4,1.732)(2,1.732)
\psdots(1,0)(-1,0)(0,1.732)(0,-1.732)(2,0)(-2,0)(4,0)(-4,0)
\psset{linewidth=1.6pt,linestyle=solid}
    \psplot{-2}2{1 x x mul 4 div sub 3 mul sqrt}
    \psplot{-2}2{1 x x mul 4 div sub 3 mul sqrt neg}
\psline(4,-2)(4,2)
\psline(-4,-2)(-4,2)
\uput[d](1,0){$\sqrt{b^2-a^2}$}
\uput[ur](2,0){$a$}
\uput[d](-1,0){$-\sqrt{b^2-a^2}$}
\uput[ul](-2,0){$-a$}
\uput[dl](0,1.732){$b$}
\uput[ul](0,-1.732){$-b$}
\uput[dl](4,0){$\frac a{\sqrt{a^2-b^2}}$}
\uput[dr](-4,0){$\frac{-a}{\sqrt{a^2-b^2}}$}
  \end{pspicture}
  \caption{Elipsin odaklar\i\ ve do\u grultmanlar\i}\label{fig:ell-foc}
\end{figure}
\end{definition}

\begin{remark}
\begin{figure}[ht]
\mbox{}\hfill
\psset{unit=2cm,plotpoints=200}
  \begin{pspicture}(-0,-0.25)(2,1)
\uput[d](0,0){$A$}
\uput[d](1,0){$B$}
\uput[d](1.4142,0){$C$}
\uput[d](0.7071,0){$D$}
\psline(0,0)(1.4142,0)
\psdots(1.4142,0)(0,0)
\psset{linewidth=1.6pt}
\psline(0.7071,0)(0.7071,1)
    \psplot1{1.4142}{1 x x mul sub neg sqrt}
  \end{pspicture}  
\hfill
  \begin{pspicture}(-2.3094,-0.25)(0,1)
\uput[d](0,0){$A$}
\uput[d](-2,0){$B$}
\uput[d](-1.7320,0){$C$}
\uput[d](-2.3094,0){$D$}
\psline(-2.3094,0)(0,0)
\psdots(-1.7320,0)(0,0)
\psset{linewidth=1.6pt}
    \psplot{-2}0{1 x x mul 4 div sub sqrt}
\psline(-2.3094,0)(-2.3094,1)
  \end{pspicture}
\hfill\mbox{}
  \caption{Odak ve do\u grultman}\label{fig:foc-dir}
  
\end{figure}
\Sekilde{fig:foc-dir}ki gibi
merkezli koni kesitinin merkezi $A$, ve
(merkezin ayn\i\ taraf\i nda olan) k\"o\c sesi $B$, ve oda\u g\i\ $C$ ise,
ve do\u grultman\i, koni kesitinin eksenini $D$ noktas\i nda keserse,
tan\i ma g\"ore koni kesitinin d\i\c smerkezlili\u gi $AC:AB$, ama
\begin{equation*}
  AC:AB::AB:AD,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
  BC:BD::AC:AB,
\end{equation*}
ve sonu\c c olarak koni kesitinin d\i\c smerkezlili\u gi $BC:BD$.
\end{remark}


\begin{theorem}\label{thm:ecc}
  Merkezli koni kesitinin noktalar\i,
bir odak noktas\i na 
ve ona kar\c s\i l\i k gelen do\u grultman do\u grusuna 
uzakl\i klar\i n\i n oran\i n\i n d\i\c smerkezlilik oldu\u gu noktalard\i r.
Yani 
%koni kesitinin d\i\c smerkezlili\u gi $\epsilon$ ise,
\c Sekil \ref{fig:ecc-hyp}
\begin{figure}
  \centering
\psset{unit=1.8cm,plotpoints=200}
  \begin{pspicture}(-2,-1.414)(2,1.414)
\psline(-1.4142,0)(1.4142,0)
\psset{linestyle=dotted}
\psline(-1.4142,0)(-1.5,1.1180)(0.7071,1.1180)
\psline(-1.5,1.1180)(1.4142,0)
\psdots(1.4142,0)(-1.4142,0)(0,0)
\psset{linestyle=solid,linewidth=1.6pt}
\psline(0.7071,-1.414)(0.7071,1.414)
\psline(-0.7071,-1.414)(-0.7071,1.414)
    \psplot1{1.732}{1 x x mul sub neg sqrt}
    \psplot1{1.732}{1 x x mul sub neg sqrt neg}
    \psplot{-1}{-1.732}{1 x x mul sub neg sqrt}
    \psplot{-1}{-1.732}{1 x x mul sub neg sqrt neg}
\uput[d](0,0){$A$}
\uput[dl](-1,0){$B$}
\uput[dr](1,0){$B'$}
\uput[l](-1.4142,0){$C$}
\uput[r](1.4142,0){$C'$}
\uput[dr](-0.7071,0){$D$}
\uput[dl](0.7071,0){$D'$}
\uput[dl](-1.5,1.1180){$E$}
\uput[ul](-0.7071,1.1180){$F$}
\uput[r](0.7071,1.1180){$F'$}
  \end{pspicture}  
  \caption{Hiperbol\"un d\i\c smerkezlilik}\label{fig:ecc-hyp}
\end{figure}
ve \numarada{fig:ecc-ell}
\begin{figure}
  \centering
\psset{plotpoints=200}
\psset{unit=1.2cm}
  \begin{pspicture}(-2,-2)(2,2)
\psline(-2.8284,0)(2.8284,0)
\psset{linestyle=dotted}%](-2,-1.4142)(2,-1.4141)(2,1.4142)(-2,1.4142)
\psline(-1.4141,0)(1,1.2247)(1.4142,0)
\psline(-2.8284,1.2247)(2.8284,1.2247)
\psdots(1.4142,0)(-1.4142,0)(0,0)
\psset{linewidth=1.6pt,linestyle=solid}
    \psplot{-2}2{1 x x mul 4 div sub 2 mul sqrt}
    \psplot{-2}2{1 x x mul 4 div sub 2 mul sqrt neg}
\psline(2.8284,-1.4142)(2.8284,1.4142)
\psline(-2.8284,-1.4142)(-2.8284,1.4142)
\uput[d](0,0){$A$}
\uput[dl](-2,0){$B$}
\uput[dr](2,0){$B'$}
\uput[d](-1.4142,0){$C$}
\uput[d](1.4142,0){$C'$}
\uput[l](-2.8284,0){$D$}
\uput[r](2.8284,0){$D'$}
\uput[ur](1,1.2247){$E$}
\uput[l](-2.8284,1.2247){$F$}
\uput[r](2.8284,1.2247){$F'$}
  \end{pspicture}
  \caption{Elipsin d\i\c smerkezlilik}\label{fig:ecc-ell}
\end{figure}
($CB=C'B'$ ve $BD=B'D'$ oldu\u gundan)
a\c sa\u g\i daki ko\c sullar denktir:
\begin{itemize}
\item 
$E$ noktas\i\ koni kesitinde,
\item
$CE:EF::CB:BD$,
\item
$C'E:EF'::CB:BD$.
\end{itemize}
\end{theorem}

\begin{remark}
$BC:BD::AB:AD::BB':DD'$
oldu\u gundan
merkezli koni kesitinin $E$ noktalar\i\ i\c cin
(ve sadece bu noktalar i\c cin)
\begin{equation*}
  C'E\pm CE:EF'\pm EF::BB':DD'.
\end{equation*}
Elipste $EF'+EF=FF'=DD'$,
dolay\i s\i yla
\begin{equation*}
  CE+C'E=BB'.
\end{equation*}
Hiperbolde $E$, soldaki daldaysa $EF'-EF=DD'$, dolay\i s\i yla
\begin{equation*}
  C'E-CE=BB'.
\end{equation*}
\end{remark}

\section{Kutupsal koordinatlar}

\begin{definition}
\Sayfada{fig:cos}ki \Sekilde{fig:cos} $\angle BAC$ dik ise
\begin{align*}
  \sin\alpha&=\frac ab,&\tan\alpha&=\frac ac,&\sec\alpha&=\frac bc,\\
  \cos\alpha&=\frac cb,&\cot\alpha&=\frac ca,&\csc\alpha&=\frac ba.
\end{align*}
\end{definition}

\begin{remark}
  \Sekilde{fig:trig}ki
  \begin{figure}
    \centering\psset{unit=1.6cm}
    \begin{pspicture}(-1,-1)(1,1)
\pscircle(0,0)1
      \pspolygon(0,0)(1,0)(1,0.75)
\psline(0.8,-0.6)(0.8,0.6)
\uput[d](0,0){$O$}
\uput[r](1,0){$A$}
\uput[r](1,0.75){$B$}
\uput[dl](0.8,0){$C$}
\uput[u](0.8,0.6){$D$}
\uput[dr](0.8,-0.6){$E$}
\uput{10pt}[18](0,0){$\theta$}
    \end{pspicture}
    \caption{Trigonometri (\"u\c cgen \"ol\c cmesi)}\label{fig:trig}
    
  \end{figure}
\c cemberin yar\i\c cap\i\ birim ise
\begin{align*}
  \sin\theta&=CD=\frac12DE,&\tan\theta&=AB,&\sec\theta&=OB.
\end{align*}
Latince'de
\begin{compactitem}
\item 
\emph{tangens, tangent-,} ``dokunan, te\u get'' demektir;
\item
\emph{secans, secant-,} ``kesen'' demektir;
\item
\emph{sinus,} ``koy, k\"orfez'' demektir.
\end{compactitem}
Latince \emph{sinus}'un matematiksel kullan\i l\i\c s\i,
Arap\c ca'dan yanl\i\c s \c ceviridir.
Arap\c ca'da\nocite{Boyer}
\begin{compactitem}
\item 
\emph{cayb,} ``koy, k\"orfez'' demektir;
\item
\emph{ciba,} ``sin\"us'' demektir.
\end{compactitem}
\end{remark}

\begin{definition}
  \Sekilde{fig:circ}
  \begin{figure}
    \centering\psset{unit=1.6cm}
    \begin{pspicture}(-1.5,-1.2)(1.5,1.2)
\pscircle[linestyle=dotted](0,0)1
\psline(-0.6,-0.8)(0.6,0.8)
\psline(-0.8,0.6)(0.8,-0.6)
\psset{linewidth=1.6pt}
      \psline{->}(-1.5,0)(1.5,0)
      \psline{->}(0,-1.2)(0,1.2)
\uput[d](1.5,0){$x$}
\uput[l](0,1.2){$y$}
\uput{10pt}[116](0,0){$O$}
%\rput[dl](0,0){\psframebox[fillstyle=solid,linecolor=white]{$O$}}
\uput[ur](1,0){$A$}
\uput[53](0.6,0.8){$B$}
\uput[143](-0.8,0.6){$C$}
\uput[233](-0.6,-0.8){$D$}
\uput[-37](0.8,-0.6){$E$}
    \end{pspicture}
    \caption{A\c c\i lar\i n \"ol\c c\"us\"u}\label{fig:circ}
    
  \end{figure}
$BOD$ ve $COE$ do\u grular\i\ birbirine dik ise ve
$\angle AOB=\alpha$ ise
  \begin{align*}
    \angle AOC&=\alpha+\frac{\uppi}2,&
\angle AOD&=\alpha+\uppi,&
\angle AOE&=\alpha+\frac{3\uppi}2.
  \end{align*}
Herhangi $\beta$ a\c c\i\ \"ol\c c\"us\"u i\c cin
\begin{equation*}
  \beta=\beta\pm 2\uppi=\beta\pm4\uppi=\dotsb
\end{equation*}
D\"uzlemde $O$ olmayan herhangi $F$ noktas\i\ i\c cin 
$OF=r$ ve $\angle AOF=\theta$ ise
$F$ noktas\i n\i n \textbf{kutupsal koordinatlar\i}
\begin{equation*}
  (r,\theta)\qquad\text{veya}\qquad(-r,\theta\pm\uppi).
\end{equation*}
$F$ noktas\i n\i n dik koordinatlar\i\ $(x,y)$ ise
\begin{align*}
  \sin\theta&=\frac yr,&\tan\theta&=\frac yx,&\sec\theta&=\frac rx,\\
\cos\theta&=\frac xr,&\cot\theta&=\frac xy,&\csc\theta&=\frac ry.
\end{align*}
Bir e\u grinin noktalar\i n\i n kutupsal koordinatlar\i n\i n 
sa\u glad\i\u g\i\ bir denklem,
e\u grinin \textbf{kutupsal denklemidir.}
\end{definition}

\begin{theorem}
D\"uzlemde $O$ olmayan bir noktan\i n dik koordinatlar\i\ $(x,y)$
ve kutupsal koordinatlar\i\ $(r,\theta)$ ise
\begin{align*}
       r^2&=x^2+y^2,& x&=r\cos\theta,\\
\tan\theta&=\frac yx,&y&=r\sin\theta.
\end{align*}
\end{theorem}

\begin{theorem}
\c Cember olmayan,
oda\u g\i\ $(0,0)$ olan,
do\u grultman\i\ $x+p=0$ olan,
d\i\c smerkezli\u gi $e$ olan, 
koni kesitinin kutupsal denklemi
\begin{equation}\label{eqn:polar}
  r=\frac{ep}{1-e\cos\theta}.
\end{equation}
(\Sekle{fig:polar} 
\begin{figure}
  \centering\psset{unit=8mm}
  \begin{pspicture}(-2,-1)(3,4.5)
\parametricplot[linewidth=1.6pt]{55}{240}{2 1 t cos sub div t cos mul
                         2 1 t cos sub div t sin mul}
\psline{->}(-3,0)(4,0)
\psline{->}(0,-1)(0,4.5)
\psline(-2,-1)(-2,4)
\psline(0,0)(2,3.464)(-2,3.464)
%\psline(2,0)(2,3.464)
\uput[u](-1,3.464){$p$}
\uput[u](1,3.464){$r\cos\theta$}
\uput[30](0,0){$\theta$}
\uput[-30](1,1.732){$r$}
\uput[r](2,2){$e=\displaystyle\frac r{p+r\cos\theta}$}
\psdots(0,0)(2,3.464)
      \end{pspicture}
  \caption{Koni kesitinin kutupsal denklemi}\label{fig:polar}
  
\end{figure}
bak\i n.)
Kar\c s\i l\i k gelen dik denklem,
\begin{equation*}
  (1-e^2)\cdot x^2+y^2=2e^2px+e^2p^2.
\end{equation*}
\end{theorem}

\begin{remark}
  $e=1/p$ durumunda \eqref{eqn:polar} denklemi
  \begin{equation*}
    r=\frac1{1-e\cos\theta}
  \end{equation*}
olur. 
Baz\i\ durumlar \Sekilde{fig:ecc-pol}
\begin{figure}
  \centering\psset{unit=9mm,labelsep=1pt}
  \begin{pspicture}*(-5.5,-4)(5.5,4)
%\psgrid
\rput(-1.5,-3){$r=\displaystyle\frac1{1-e\cos\theta}$}
\pscircle(0,0)1
\parametricplot{ 0}{360}{1 1 t cos 0.512 mul sub div t cos mul
                         1 1 t cos 0.512 mul sub div t sin mul}
\parametricplot{ 0}{360}{1 1 t cos 0.64 mul sub div t cos mul
                         1 1 t cos 0.64 mul sub div t sin mul}
\parametricplot{ 0}{360}{1 1 t cos 0.8 mul sub div t cos mul
                         1 1 t cos 0.8 mul sub div t sin mul}
\parametricplot{10}{350}{1 1 t cos sub div t cos mul
                         1 1 t cos sub div t sin mul}
\parametricplot{37}{323}{1 1 t cos 1.25 mul sub div t cos mul
                         1 1 t cos 1.25 mul sub div t sin mul}
\parametricplot{-36}{36}{1 1 t cos 1.25 mul sub div t cos mul
                         1 1 t cos 1.25 mul sub div t sin mul}
\parametricplot{51}{309}{1 1 t cos 1.5625 mul sub div t cos mul
                         1 1 t cos 1.5625 mul sub div t sin mul}
\parametricplot{-50}{50}{1 1 t cos 1.5625 mul sub div t cos mul
                         1 1 t cos 1.5625 mul sub div t sin mul}
\psline{->}(-5.5,0)(5.5,0)
\psline{->}(0,-4)(0,4)
\uput[ul](-0.8,0.6){$e=0$}
%\uput[dl](1.8,0.6){$e=64/125$}
\uput[ur](2.8,0){$e=16/25$}
\uput[ur](4.2,1.2){$e=4/5$}
\uput[dr](4,3){$e=1$}
\uput[ur](-4.4,1){$e=5/4$}
\uput[ur](-3.6,3){$e=25/16$}
  \end{pspicture}
  \caption{D\i\c smerkezli\u ge g\"ore koni kesitleri}\label{fig:ecc-pol}
  
\end{figure}
g\"or\"un\"ur.
\end{remark}

\begin{theorem}(\Sekle{fig:proj}
  \begin{figure}
    \centering\psset{unit=2cm}
    \begin{pspicture}(-1.5,-1.1)(1.5,1.4)
      \pscircle(0,0)1
\pscircle(0.5,0){0.5}
\psline(1,-1.1)(1,1.4)
\psline{->}(-1.5,0)(1.5,0)
\psline{->}(0,-1.1)(0,1.4)
\psline(0,0)(1,1.333)
\uput[ul](-0.707,0.707){$r=1$}
\uput{1pt}[ur](0.15,-0.35){$r=\cos\theta$}
\uput[r](1,0.6){$r=\sec\theta$}
\uput[dl](0,0){$O$}
\uput[ul](0.36,0.48){$A$}
\uput[r](0.6,0.8){$B$}
\uput[r](1,1.333){$C$}
    \end{pspicture}
    \caption{\c Cemberler ve do\u gru}\label{fig:proj}
    
  \end{figure}
bak\i n.)
  \begin{itemize}
  \item 
$r=\cos\theta$ kutupsal denklemi,
merkezi $(1/2,0)$ olan ve yar\i\c cap\i\ $1/2$ olan \c cemberi tan\i mlar.
  \item 
$r=\sin\theta$ kutupsal denklemi,
merkezi $(0,1/2)$ olan ve yar\i\c cap\i\ $1/2$ olan \c cemberi tan\i mlar.
\item
$r=\sec\theta$ kutupsal denklemi,
$x=1$ do\u grusunu tan\i mlar.
\item
$r=\csc\theta$ kutupsal denklemi,
$y=1$ do\u grusunu tan\i mlar.
  \end{itemize}
\end{theorem}

\begin{remark}
\Sekilde{fig:proj} $OA:OB::OB:OC$, ama $OB=1$, dolay\i s\i yla
\begin{equation*}
OA\cdot OC=1.
\end{equation*}
\end{remark}

\begin{theorem}
  (\c Sekiller \ref{fig:rose2},
  \begin{figure}
    \centering
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
\psdots(1,0)
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{45}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{90}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{135}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}

    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{180}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{225}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{270}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{315}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
    \end{pspicture}

\psset{unit=3.5cm}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{360}{2 t mul cos t cos mul
                            2 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1,-1)(1,1)
\psline(1,-1)(-1,1)
\uput[dl](-1,-1){$\theta=\uppi/4$}
\uput[dr](1,-1){$\theta=3\uppi/4$}
\uput[d](0,-1){$r=\cos(2\theta)$}
\uput[l](-0.866,-0.5){$r=1$}
    \end{pspicture}
\caption{$4$-yaprakl\i\ g\"ul}\label{fig:rose2}
  \end{figure}
\ref{fig:rose3},
\begin{figure}
    \centering
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
\psdots(1,0)
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{30}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{60}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}

    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{90}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{120}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}
    \begin{pspicture}(-1.2,-1.2)(1.2,1.2)
      \parametricplot0{150}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1.2)
    \end{pspicture}

\psset{unit=3.5cm}
    \begin{pspicture}(-1.2,-1.4)(1.2,1)
      \parametricplot0{180}{3 t mul cos t cos mul
                            3 t mul cos t sin mul}
\psset{linestyle=dotted}
\pscircle(0,0)1
\psline(-1.039,-0.6)(1.039,0.6)
\psline(-1.039,0.6)(1.039,-0.6)
\psline(0,-1.2)(0,1)
\uput[30](1.039,0.6){$\theta=\uppi/6$}
\uput[-30](1.039,-0.6){$\theta=-\uppi/6$}
\uput[d](0,-1.2){$\theta=\uppi/2$}
\uput[l](-1,0){$r=1$}
\uput[d](-0.5,-0.866){$r=\cos(3\theta)$\phantom{$r=\cos(3\theta)$}}
    \end{pspicture}
    \caption{$3$-yaprakl\i\ g\"ul}\label{fig:rose3}
    
  \end{figure}
\ref{fig:rose45},
\begin{figure}
\mbox{}\hfill\psset{unit=2.4cm}
    \begin{pspicture}(-1,-1.2)(1,1)
      \parametricplot0{360}{4 t mul cos t cos mul
                            4 t mul cos t sin mul}
\uput[d](0,-1){$r=\cos(4\theta)$}
    \end{pspicture}
\hfill
    \begin{pspicture}(-1,-1.2)(1,1)
      \parametricplot0{180}{5 t mul cos t cos mul
                            5 t mul cos t sin mul}
\uput[d](0,-1){$r=\cos(5\theta)$}
    \end{pspicture}
\hfill\mbox{}

  \caption{$8$- ve $5$-yaprakl\i\ g\"uller}\label{fig:rose45}
  
\end{figure}
ve
\numaraya{fig:roses}
\begin{figure}
  \centering\psset{unit=12mm}
    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{t sin t cos mul
                            t sin t sin mul}
\uput[d](0,-1){$r=\sin\theta$}
    \end{pspicture}
    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{2 t mul sin t cos mul
                            2 t mul sin t sin mul}
\uput[d](0,-1){$r=\sin(2\theta)$}
    \end{pspicture}
    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{3 t mul sin t cos mul
                            3 t mul sin t sin mul}
\uput[d](0,-1){$r=\sin(3\theta)$}
    \end{pspicture}

    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{4 t mul sin t cos mul
                            4 t mul sin t sin mul}
\uput[d](0,-1){$r=\sin(4\theta)$}
    \end{pspicture}
    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{5 t mul sin t cos mul
                            5 t mul sin t sin mul}
\uput[d](0,-1){$r=\sin(5\theta)$}
    \end{pspicture}
    \begin{pspicture}(-1,-1.5)(1,1)
      \parametricplot0{360}{6 t mul sin t cos mul
                            6 t mul sin t sin mul}
\uput[d](0,-1){$r=\sin(6\theta)$}
    \end{pspicture}

  \caption{G\"uller}\label{fig:roses}
  
\end{figure}
bak\i n.)  Her $n$ do\u gal say\i s\i\ i\c cin
\begin{equation*}
  r=\cos(n\theta)\qquad\text{ve}\qquad r=\sin(n\theta)
\end{equation*}
kutupsal denklemlerinin her biri,
\begin{itemize}
\item
$n$ say\i s\i n\i n \c cift oldu\u gu durumda
\textbf{$2n$-yaprakl\i\ g\"ul\"u} tan\i mlar.
\item
$n$ say\i s\i n\i n tek oldu\u gu durumda
\textbf{$n$-yaprakl\i\ g\"ul\"u} tan\i mlar.
\end{itemize}
\end{theorem}

\begin{remark}
\Sekilde{fig:card}
\begin{figure}
    \centering\psset{unit=7mm}
    \begin{pspicture}(-0,-1.8)(1,1.8)
      \parametricplot0{360}{t cos t cos mul
                            t cos t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.02,-1.8)(1.25,1.8)
      \parametricplot0{360}{t cos 0.25 add t cos mul
                            t cos 0.25 add t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.06,-1.8)(1.5,1.8)
      \parametricplot0{360}{t cos 0.5 add t cos mul
                            t cos 0.5 add t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.14,-1.8)(1.75,1.8)
      \parametricplot0{360}{t cos 0.75 add t cos mul
                            t cos 0.75 add t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.25,-1.8)(2,1.8)
      \parametricplot0{360}{t cos 1.0 add t cos mul
                            t cos 1.0 add t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.39,-1.8)(2.25,1.8)
      \parametricplot0{360}{t cos 1.25 add t cos mul
                            t cos 1.25 add t sin mul}
    \end{pspicture}
    \begin{pspicture}(-0.56,-1.8)(2.5,1.8)
      \parametricplot0{360}{t cos 1.5 add t cos mul
                            t cos 1.5 add t sin mul}
    \end{pspicture}

\psset{unit=2.7cm}    
\begin{pspicture}(-0.6,-1.8)(2.5,1.8)
%\psgrid
      \parametricplot0{360}{t cos 0.0 add t cos mul
                            t cos 0.0 add t sin mul}
      \parametricplot0{360}{t cos 0.25 add t cos mul
                            t cos 0.25 add t sin mul}
      \parametricplot0{360}{t cos 0.5 add t cos mul
                            t cos 0.5 add t sin mul}
      \parametricplot0{360}{t cos 0.75 add t cos mul
                            t cos 0.75 add t sin mul}
      \parametricplot0{360}{t cos 1.0 add t cos mul
                            t cos 1.0 add t sin mul}
      \parametricplot0{360}{t cos 1.25 add t cos mul
                            t cos 1.25 add t sin mul}
      \parametricplot0{360}{t cos 1.5 add t cos mul
                            t cos 1.5 add t sin mul}
    \end{pspicture}
    \caption{Limasonlar}\label{fig:card}
    
  \end{figure}
g\"or\"unen e\u griler,
$a\in\left\{0,\frac14,\frac12,\frac34,1,\frac54,\frac32\right\}$ durumlar\i nda
\begin{equation*}
  r=a+\cos\theta
\end{equation*}
kutupsal denklemi taraf\i ndan tan\i mlan\i r.
Bu e\u grilerin her birine \textbf{limason} denebilir
(Frans\i zca'da \emph{lima\c con,} salyangoz demektir);\nocite{Vygodsky}
$a=1$ durumunda e\u gri \textbf{kardiyoid}\nocite{LSJ}
(\gr{kardioeid'hs}, ``kalp \c sekli'' demektir).
\end{remark}

\begin{theorem}\label{thm:lem}
  $(\pm1,0)$ noktalar\i na uzakl\i klar\i n\i n \c carp\i m\i\ birim olan
noktalar\i n yeri,
\begin{equation*}
  r^2=2\cos(2\theta)
\end{equation*}
kutupsal denklemi taraf\i ndan tan\i mlan\i r.
(\Sekle{fig:lem}
\begin{figure}
  \centering\psset{unit=3.6cm}
  \begin{pspicture}(-1.414,-0.5)(1.414,0.5)
%\psgrid
    \parametricplot{-45}{ 45}{2 2 t mul cos mul sqrt t cos mul
                              2 2 t mul cos mul sqrt t sin mul}
    \parametricplot{135}{225}{2 2 t mul cos mul sqrt t cos mul
                              2 2 t mul cos mul sqrt t sin mul}
\psdots(-1,0)(1,0)
  \end{pspicture}
  \caption{Lemniskat}\label{fig:lem}
  
\end{figure}
bak\i n.)
\end{theorem}

\begin{remark}
\Teoremde{thm:lem} tan\i mlanan e\u griye \textbf{lemniskat} denir.
Eski Yu\-nan\-ca \gr{lhmn'iskos}, ``\c serit, kurdele''\nocite{Celgin}
demektir.
(\c Ca\u gda\c s Yunanca \gr{kord'el\-la} vard\i r;\nocite{POGD} 
\.Italyanca
\emph{cordola} vard\i r.\nocite{Nis-Et}
Bunlar Eski Yunanca \gr{qord'h} kelimesinden gelir.)
\end{remark}

\section{Uzay}

\begin{definition}
Tan\i m \numarada{def:axes}ki gibi
$xy$ eksenleri verilmi\c s ise,
onlar\i n $O$ kesi\c sim noktas\i ndan ge\c cen
ve onlar\i n d\"uzleminde olmayan \textbf{$z$ ekseni} eklenebilir.
\end{definition}

\begin{theorem}\label{thm:coord3}
  $Xyz$ eksenleriyle donat\i lm\i\c s uzayda,
\Teoremde{thm:coord}ki gibi 
her $A$ noktas\i\ i\c cin
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
$z$ ekseninde bir ve tek bir $D$ i\c cin
\begin{compactitem}
  \item
$AB$ do\u grusu, $yz$ d\"uzlemine paraleldir,
  \item
$AC$ do\u grusu, $xz$ d\"uzlemine paraleldir,
  \item
$AD$ do\u grusu, $xy$ d\"uzlemine paraleldir.%%%%%
\footnote{Dejenere durumda $A$ zaten bir eksendedir.}
%%%%%%%%%%%%%%%%%%
\end{compactitem}
Tam tersine $b$, $c$, ve $d$ i\c saretli uzunluk olmak \"uzere, 
herhangi bir $(b,c,d)$ s\i ral\i\ \"u\c cl\"us\"u i\c cin,
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
d\"uzlemde bir ve tek bir $A$ i\c cin
\begin{align*}
  \ydp{OB}&=b,&\ydp{OC}&=c,&\ydp{OD}&=d,
\end{align*}
ve yukar\i daki paralellik ko\c sullar\i\ sa\u glan\i r.
\end{theorem}

\begin{definition}
  \Teoremde{thm:coord3} 
$b$, $A$ noktas\i n\i n \textbf{$x$ koordinat\i d\i r;}
$c$, $A$ noktas\i n\i n \textbf{$y$ koordinat\i d\i r;}
ve $d$, $A$ noktas\i n\i n \textbf{$z$ koordinat\i d\i r.}
``Koordinatlar\i\ $(b,c,d)$ olan nokta'' ifadesinin yerine
``$(b,c,d)$ noktas\i'' diyebiliriz.
\end{definition}

\begin{theorem}\label{thm:dist3}\sloppy
  $Xyz$ eksenleri dik ise $(a,b,c)$ noktas\i n\i n 
$(d,e,f)$ noktas\i ndan uzakl\i\u g\i
  \begin{equation*}
    \sqrt{(a-d)^2+(b-e)^2+(e-f)^2}.
  \end{equation*}
\end{theorem}

\begin{theorem}\label{thm:cone3}
E\u ger
\begin{align*}
  a^2+b^2&\neq0,&c&\neq0,&d&\neq0
\end{align*}
ise,
o zaman dik $xyz$ eksenlerine g\"ore
  \begin{equation*}
    y^2+z^2=d^2x^2
  \end{equation*}
denklemi, bir dik koninin y\"uzeyini tan\i mlar, ve
\begin{equation*}
  ax+by=c
\end{equation*}
denklemi, koniyi kesen bir d\"uzlemi tan\i mlar.
Bu \c sekilde bir koni kesiti belirtilir.
\end{theorem}

\begin{remark}
  \Teoremi{thm:dist3} kullanarak
\Teoremde{thm:cone3}ki koni kesitinin \"ozelliklerini kan\i tlanabilir.
\"Ozel olarak kesitin rastgele noktas\i n\i n ordinat\i n\i n
ve kar\c s\i l\i k gelen absisin uzunluklar\i\ bulunabilir
ve onlar\i n ili\c skisi belirtilebilir.
Ama \Teoremde{thm:cone3}ki koni diktir,
ve Teorem \ref{thm:para} ve \numarada{thm:hyp}ki koniler dik olmayabilir.
\end{remark}

\section{Vekt\"orler}

\begin{remark}
  Tan\i m \numarada{def:vec} \emph{vekt\"orler} tan\i mlad\i k.
Bu tan\i m, uzayda da ge\c cerlidir.
\end{remark}

\begin{theorem}\label{thm:ABOC}
  Ya $xy$ d\"uzleminde ya da $xyz$ uzay\i nda
her $\ydp{AB}$ y\"onl\"u do\u grusu i\c cin,
bir ve tek bir $C$ noktas\i\ i\c cin
\begin{equation*}
  \ydp{AB}=\ydp{OC}.
\end{equation*}
\end{theorem}

\begin{definition}\sloppy
\Teorem{thm:ABOC} sayesinde
``$\ydp{OC}$ vekt\"or\"u'' ifadesinin yerine
``$C$ vekt\"or\"u'' diyebiliriz.
Ayr\i ca bir vekt\"or i\c cin
\begin{equation*}
  \vec c
\end{equation*}
gibi bir ifade kullan\i labilir: 
d\"uzlemde $\vec c$ bir $(c_1,c_2)$ noktas\i n\i\ belirtir;
uzayda bir $(c_1,c_2,c_3)$ noktas\i n\i\ belirtir.
Vekt\"orler toplanabilir ve \c co\u galt\i labilir:
\begin{align*}
(a_1,a_2,a_3)+(b_1,b_2,b_3)&=(a_1+b_1,a_2+b_2,a_3+b_3),\\
a\cdot(b_1,b_2,b_3)&=(ab_1,ab_2,ab_3).
\end{align*}
Ayr\i ca birbirini \c carpabilir,
ama sonu\c c bir uzunluk, yani bir \textbf{skalerdir:}
\begin{equation*}
  (a_1,a_2,a_3)\cdot(b_1,b_2,b_3)=a_1b_1+a_2b_2+a_3b_3.
\end{equation*}
D\"uzlemde benzer tan\i mlar kullan\i l\i r.
Ya d\"uzlemde ya da uzayda
\begin{equation*}
\norm{\vec a}=\sqrt{\vec a\cdot\vec a}.
%  \norm{(a_1,a_2,a_3)}=\sqrt{a_1{}^2+a_2{}^2+a_3{}^2}.
\end{equation*}
\.Iki vekt\"or\"un \textbf{a\c c\i s\i} vard\i r:
nokta olarak vekt\"orler $A$ ve $B$ ise
 vekt\"orlerin a\c c\i s\i
 \begin{equation*}
   \angle AOB.
 \end{equation*}
E\u ger $\vec a$ ve $\vec b$ vekt\"orleri birbirine dik ise
(yani a\c c\i s\i\ $\uppi/2$ ise),
\begin{equation*}
  \vec a\perp\vec b
\end{equation*}
ifadesi kullan\i labilir.
Vekt\"orler paralel ise (yani a\c c\i s\i\ $0$ veya $\uppi$ ise)
\begin{equation*}
  \vec a\parallel\vec b
\end{equation*}
ifadesi kullan\i labilir.
\end{definition}

\begin{theorem}\label{thm:perp}
Dik eksenlere g\"ore
bir $\vec a$ vekt\"or\"un\"un uzunlu\u gu
  \begin{equation*}
    \norm{\vec a},
  \end{equation*}
ve $\vec a$ ve $\vec b$ vekt\"orlerinin a\c c\i s\i\ $\theta$ ise
\begin{equation*}
  \vec a\cdot\vec b=\norm{\vec a}\cdot\norm{\vec b}\cdot\cos\theta.
\end{equation*}
\"Ozel olarak
\begin{equation*}
  \vec a\perp\vec b\iff\vec a\cdot\vec b=0.
\end{equation*}
\end{theorem}

\begin{remark}
Teoremin kan\i t\i, Kosin\"us Teoremini (yani \Teoremi{thm:cos})
kullan\i r.
Soyut bir ``i\c c \c carp\i m uzay\i nda''
\Teorem{thm:perp}, uzunluklar\i n ve a\c c\i lar\i n \emph{tan\i m\i d\i r.}
\end{remark}

\begin{theorem}
  Dik $xyz$ eksenleriyle donat\i lm\i\c s uzayda,
s\i f\i r olmayan bir $\vec a$ vekt\"or\"une dik olan
ve bir $\vec b$ noktas\i ndan ge\c cen d\"uzleminin denklemi
\begin{equation*}
  \vec a\cdot(x,y,z)=\vec a\cdot\vec b.
\end{equation*}
\end{theorem}

\begin{remark}
  Uzayda bir d\"uzlemin denklemi,
$a^2+b^2+c^2\neq0$ olmak \"uzere
  \begin{equation*}
(a,b,c)\cdot(x,y,z)=d\qquad\text{veya}\qquad
    ax+by+cz=d
  \end{equation*}
bi\c ciminde yaz\i labilir.
\end{remark}

\begin{theorem}
  Dik $xyz$ eksenleriyle donat\i lm\i\c s uzayda,
bir $\vec c$ noktas\i n\i n
bir $\vec a\cdot(x,y,z)=\vec a\cdot\vec b$ d\"uzlemine uzakl\i\u g\i
\begin{equation*}
  \frac{\abs{\vec a\cdot(\vec c-\vec b)}}{\norm{\vec a}}.
\end{equation*}
\end{theorem}

\begin{proof}
E\u ger $\vec b-\vec c$ ve $\vec a$ vekt\"orlerinin a\c c\i s\i\ $\theta$ ise
istedi\u gimiz uzakl\i k
\begin{equation*}
  \norm{\vec c-\vec b}\cdot\abs{\cos\theta},
\end{equation*}
yani (\Teoreme{thm:perp} g\"ore)
\begin{gather*}
  \frac{\abs{\vec a\cdot(\vec c-\vec b)}}{\norm{\vec a}}.\qedhere
\end{gather*}
\end{proof}

\begin{remark}
  Teoremde d\"uzlemin denklemi $ax+by+cz=d$ ve nokta $(s,t,u)$ ise
istenen uzakl\i k
\begin{equation*}
  \frac{\abs{as+bt+cu-d}}{\sqrt{a^2+b^2+c^2}}.
\end{equation*}
\end{remark}

\begin{remark}
E\u ger $\vec a$, $\vec b$, ve $\vec c$ noktalar\i\ verilirse,
s\i f\i r olmayan bir $\vec d$ i\c cin
verilmi\c s noktalardan ge\c cen d\"uzlem
\begin{equation*}
  \vec d\cdot(x,y,z)=\vec d\cdot\vec a\qquad\text{veya}\qquad
\vec d\cdot\bigl((x,y,z)-\vec a\bigr)=0
\end{equation*}
bi\c ciminde yaz\i labilir.
O zaman $\vec d$ vekt\"or\"u,
\begin{equation}\label{eqn:plane}
  \left\{
  \begin{gathered}
    (\vec b-\vec a)\cdot(x,y,z)=0\\
    (\vec c-\vec a)\cdot(x,y,z)=0
  \end{gathered}
\right.
\end{equation}
yani
\begin{equation*}
  \left\{
  \begin{gathered}
(b_1-a_1)\cdot x+(b_2-a_2)\cdot y+(b_3-a_3)\cdot z=0\\
(c_1-a_1)\cdot x+(c_2-a_2)\cdot y+(c_3-a_3)\cdot z=0
  \end{gathered}
\right.
\end{equation*}
do\u grusal denklem sisteminin s\i f\i r olmayan bir \c c\"oz\"um\"ud\"ur.
\end{remark}

\begin{definition}
Uzayda
\begin{equation*}
  \vec a\times\vec b
=\left(
\begin{vmatrix}
  a_2&a_3\\b_2&b_3
\end{vmatrix},
-\begin{vmatrix}
  a_1&a_3\\b_1&b_3
\end{vmatrix},
\begin{vmatrix}
  a_1&a_2\\b_1&b_2
\end{vmatrix}
\right).
\end{equation*}
Burada
  \begin{equation*}
    \begin{vmatrix}
      a&b\\c&d
    \end{vmatrix}
=ad-bc.
  \end{equation*}
\end{definition}

\begin{theorem}
  Uzayda
  \begin{align*}
    \vec a\times\vec b&\perp\vec a,&
    \vec a\times\vec b&\perp\vec b.
  \end{align*}
Ayr\i ca $\vec a$ ve $\vec b$ vekt\"orlerinin a\c c\i s\i\ $\theta$ ise
\begin{equation*}
  \norm{\vec a\times\vec b}=\norm{\vec a}\cdot\norm{\vec b}\cdot\sin\theta,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
  \vec a\parallel\vec b\iff\vec a\times\vec b=\vec0.
\end{equation*}
\end{theorem}

\begin{remark}
  Sonu\c c olarak \eqref{eqn:plane} sisteminin bir \c c\"oz\"um\"u,
  \begin{equation*}
    (\vec b-\vec a)\times(\vec c-\vec a).
  \end{equation*}
\end{remark}

\begin{remark}
  Uzayda $\vec a\cdot(x,y,z)=b$ ve $\vec c\cdot(x,y,z)=d$ 
d\"uzlemleri paraleldir (veya ayn\i d\i r) ancak ve ancak
  \begin{equation*}
    \vec a\parallel\vec c.
  \end{equation*}
D\"uzlemler parelel de\u gilse, bir do\u gruda kesi\c sir.
Bu durumda do\u gru $\vec a\times\vec c$ vekt\"or\"une paraleldir.
Do\u gru bir $\vec e$ noktas\i ndan ge\c cerse,
do\u grunun her noktas\i
\begin{equation*}
  \vec e+t\cdot(\vec a\times\vec b)
\end{equation*}
bi\c ciminde yaz\i labilir.
\c Simdi $\vec a\times\vec b=\vec f$ olsun.
Do\u grunun \textbf{parametrik denklemi}
\begin{equation*}
  (x,y,z)=\vec e+t\cdot\vec f.
\end{equation*}
Bu denklem
\begin{equation*}
  \left\{
  \begin{gathered}
    x=e_1+f_1t\\
y=e_2+f_2t\\
z=e_3+f_3t
  \end{gathered}
\right.
\end{equation*}
bi\c ciminde yaz\i labilir.
Buradan iki d\"uzlemin denklemleri \c c\i kar,
ve her birinde, de\u gi\c skenlerin biri g\"or\"unmez.
\"Orne\u gin $f_1f_2f_3\neq0$ ise
\begin{equation*}
  \frac{x-e_1}{f_1}=\frac{y-e_2}{f_2}=\frac{z-e_3}{f_3}.
\end{equation*}
\end{remark}

\nocite{Karakas}
\nocite{MR1660991}
\nocite{Apollonius-Heiberg}
\nocite{Descartes-Geometry}

%\bibliographystyle{plain}
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\end{document}
