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\begin{document}
\title{Analitik Geometri \"Ozeti}
\author{David Pierce}
\date{\today, \currenttime}
\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
%\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{equation*}
  \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{equation*}
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}
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}
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}
  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)
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\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)
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\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){$X$}
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\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}
\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 $2ay^2=2a\ell x+\ell x^2$ 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 $2ay^2=\ell x^2-\ell a^2$ 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}
  $Xyz$ 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}
\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}
%\bibliography{../references}

\begin{thebibliography}{10}

\bibitem{Apollonius-Heiberg}
{Apollonius of Perga}.
\newblock {\em Apollonii {P}ergaei Qvae {G}raece Exstant Cvm Commentariis
  Antiqvis}, volume~I.
\newblock Teubner, 1974.
\newblock Edited with Latin interpretation by I. L. Heiberg. First published
  1891.

\bibitem{MR1660991}
{Apollonius of Perga}.
\newblock {\em Conics. {B}ooks {I}--{III}}.
\newblock Green Lion Press, Santa Fe, NM, revised edition, 1998.
\newblock Translated and with a note and an appendix by R. Catesby Taliaferro,
  with a preface by Dana Densmore and William H. Donahue, an introduction by
  Harvey Flaumenhaft, and diagrams by Donahue, edited by Densmore.

\bibitem{MR2093668}
Archimedes.
\newblock {\em The Two Books \emph{{O}n the Sphere and the Cylinder}}, volume~I
  of {\em The Works of {A}rchimedes}.
\newblock Cambridge University Press, Cambridge, 2004.
\newblock Translated into English, together with Eutocius' commentaries, with
  commentary, and critical edition of the diagrams, by Reviel Netz.

\bibitem{Boyer}
Carl~B. Boyer.
\newblock {\em A History of Mathematics}.
\newblock John Wiley \& Sons, New York, 1968.

\bibitem{Celgin}
G{\"u}ler {\c C}elgin.
\newblock {\em Eski {Y}unanca--{T}{\"u}rk{\c c}e S{\"o}zl{\"u}k}.
\newblock Kabalc{\i}, {\.I}stanbul, 2011.

\bibitem{Descartes-Geometry}
Ren{\'e} Descartes.
\newblock {\em The Geometry of {R}en{\'e} {D}escartes}.
\newblock Dover Publications, Inc., New York, 1954.
\newblock Translated from the French and Latin by David Eugene Smith and Marcia
  L. Latham, with a facsimile of the first edition of 1637.

\bibitem{Karakas}
H.~I. Karaka{\c{s}}.
\newblock {\em Analytic Geometry}.
\newblock M $\oplus$ V [Matematik Vakf{\i}], [Ankara], n.d. [1994].

\bibitem{LSJ}
Henry~George Liddell and Robert Scott.
\newblock {\em A {G}reek-{E}nglish Lexicon}.
\newblock Clarendon Press, Oxford, 1996.
\newblock Revised and augmented throughout by Sir Henry Stuart Jones, with the
  assistance of Roderick McKenzie and with the cooperation of many scholars.
  With a revised supplement.

\bibitem{Nis-Et}
Sevan Ni{\c s}anyan.
\newblock {\em S{\"o}zlerin Soya{\u g}ac{\i}: {\c C}a{\u g}da{\c s}
  {T}{\"u}rk{\c c}enin Etimolojik S{\"o}zl{\"u}{\u g}{\"u}}.
\newblock Adam Yay{\i}nlar{\i}, {\.I}stanbul, 3rd edition, 2007.
\newblock ``The Family Tree of Words: An Etymological Dictionary of
  Contemporary Turkish.'' Geni{\c s}letilmi{\c s} g{\"o}zden ge{\c c}irilmi{\c
  s} (``expanded and revised'').

\bibitem{Oklid-2014-T}
{\"O}klid.
\newblock {\em {\"O}{\u g}elerin 13 Kitab{\i}ndan Birinci Kitap}.
\newblock Matematik B{\"o}l{\"u}m{\"u}, Mimar Sinan G{\"u}zel Sanatlar
  {\"U}niversitesi, {\.I}stanbul, 4th edition, Eyl{\"u}l 2014.
\newblock {\"O}klid'in Yunanca metni ve {\"O}zer {\"O}zt{\"u}rk \&\ David
  Pierce'in {\c c}evirdi{\u g}i T{\"u}rk{\c c}esi.

\bibitem{POGD}
J.~T. Pring, editor.
\newblock {\em The Pocket {O}xford {G}reek Dictionary}.
\newblock Oxford University Press, 2000.

\bibitem{MR13:419a}
Ivor Thomas, editor.
\newblock {\em Selections Illustrating the History of {G}reek Mathematics.
  {V}ol. {I}. {F}rom {T}hales to {E}uclid}.
\newblock Number 335 in Loeb Classical Library. Harvard University Press,
  Cambridge, Mass., 1951.
\newblock With an English translation by the editor.

\bibitem{Vygodsky}
M.~Vygodsky.
\newblock {\em Mathematical Handbook: {H}igher Mathematics}.
\newblock Mir Publishers, Moscow, 1975.
\newblock Translated from the Russian by George Yankovsky. Fifth printing 1987.

\end{thebibliography}


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