\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

\begin{center}
{\bf Coordinate Geometry}
\end{center}

{\bf Equations of Second degree}

The problem dealt with here is the following, given an expression
quadratic in $x$ and $y$, what curve does it represent in
rectangular cartesian co-ordinates.

To answer this question we shall need to be able to translate and
rotate axes in cartesian co-ordinates.

Translation

\setlength{\unitlength}{0.5in}
\begin{picture}(10,4)
\put(0,0){\vector(1,0){4}}

\put(4.2,0){\makebox(0,0){$x$}}

\put(0,0){\vector(0,1){3}}

\put(0,3.2){\makebox(0,0){$y$}}

\put(1,1){\vector(1,0){4}}

\put(5.2,1){\makebox(0,0){$X$}}

\put(1.5,0.5){\vector(0,1){3}}

\put(1.5,3.7){\makebox(0,0){$Y$}}

\put(1.5,1){\circle*{0.1}}

\put(1.9,0.8){\makebox(0,0){$(h,k)$}}

\put(3.5,2.5){\circle*{0.1}}

\put(4.1,2.7){\makebox(0,0){$P(x,y)$}}

\put(4.1,2.3){\makebox(0,0){$P(X,Y)$}}

\put(7,3){\makebox(0,0){$x=X+k$}}

\put(7,2.6){\makebox(0,0){$y=Y+k$}}

\put(7,2){\makebox(0,0){$\updownarrow$}}

\put(7,1.4){\makebox(0,0){$X=x-h$}}

\put(7,1){\makebox(0,0){$Y=y-k$}}

\end{picture}

Rotation

\setlength{\unitlength}{0.5in}
\begin{picture}(10,5)
\put(3,0.5){\vector(1,0){4.5}}

\put(7.7,0.5){\makebox(0,0){$x$}}

\put(3.5,0){\vector(0,1){3.5}}

\put(3.5,3.7){\makebox(0,0){$y$}}

\put(3.7,0.2){\makebox(0,0){$O$}}

\put(3.1,0.1){\vector(1,1){3.5}}

\put(6.8,3.8){\makebox(0,0){$X$}}

\put(3.5,0.5){\vector(-1,1){3}}

\put(0.3,3.7){\makebox(0,0){$Y$}}

\put(3.9,0.7){\makebox(0,0){$\alpha$}}

\multiput(4.5,0.5)(0,0.1){30}{\circle*{.05}}

\put(4.5,0.3){\makebox(0,0){$M$}}

\multiput(5.5,0.5)(0,0.1){20}{\circle*{.05}}

\put(5.5,0.3){\makebox(0,0){$N$}}

\put(5.7,2.4){\makebox(0,0){$Q$}}

\put(4.2,2.5){\makebox(0,0){$R$}}

\multiput(4.5,2.5)(0.1,0){11}{\circle*{.05}}

\multiput(4.5,3.5)(0.1,-0.1){10}{\circle*{.05}}

\put(4.5,3.5){\circle*{0.1}}

\put(5.2,3.8){\makebox(0,0){$P(x,y)$}}

\put(5.3,3.4){\makebox(0,0){$P(X,Y)$}}

\put(4.7,3.1){\makebox(0,0){$\alpha$}}

\end{picture}


$\ds x=OM=ON-MN=ON-QR=X\cos\alpha-Y\sin\alpha$

$\ds y=MP=MR+RP=QN+RP=X\sin\alpha+Y\cos\alpha$

$ \begin{array}{l} x=X\cos\alpha-Y\sin\alpha\\
y=X\sin\alpha+Y\cos\alpha\end{array} \longleftrightarrow
\begin {array}{l} X=x\cos\alpha+y\sin\alpha\\
Y=y\cos\alpha-x\sin\alpha\end{array}$

Using these transformations it can be shown that the following
cases occur.  The equation $\ds Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
represents

\begin{itemize}
\item[I)]
A parabola (including the degenerate case of parallel lines) if

$B^2-4AC=0$.

\item[II)]
An ellipse, or possibly no real locus, if $B^2-4AC<0$.  The
ellipse is a circle if $B=0$ and $A=C$.  It may degenerate to a
point.

\item[III)]
A hyperbola (or a pair of intersecting straight lines) if
$B^2-4AC>0$.
\end{itemize}

To reduce an equation to a standard form

\begin{itemize}
\item[I)]
rotate the axes through angle $\alpha$, to make the product term
vanish $\ds \left(\tan2\alpha=\frac{B}{A-C}\right)$

\item[II) and III)]
First translate the origin so that the terms of degree 1 vanish.
Then rotate the new axes through angle $\alpha$, to make the
product term vanish $\ds \left(\tan2\alpha=\frac{B}{A-C}\right)$
\end{itemize}

For proofs of the above assertions see D.S.Jones and D.W.Jordan
Introductory Analysis Volume 1 pp54-58.

Example
\begin{itemize}
\item[i)]
Investigate the conic $\ds 3x^2+5xy-2y^2-x+5y-2=0 \hspace{0.5in}
(1)$

$\ds B^2-4AC=25+24=49>0$

So we have a hyperbola or a pair of lines.

Put $x=X+h \hspace{0.2in} y=Y+k$

Equation (1) then becomes

$\ds 3(X+h)^2+5(X+h)(Y+k)-2(Y+k)^2-(X+h)+5(Y+k)-2=0$

$\ds 3X^2+5XY-2Y^2+X(6h+5k-1)+Y(5h-4k+5)+3h^2$

$\ds \hspace{0.35in} +5hk-2k^2-h+5k-2=0 \hspace{0.2in} (2)$

We choose $h,k$ to satisfy

$\ds 6h+5k-1=0 \hspace{0.5in} 5h-4k+5=0$

so $h=-\frac{3}{7} \hspace{1in} k=\frac{5}{7}$

Equation (2) then becomes $\ds 3X^2+5XY-2Y^2=0$

$\ds (3X-Y)(X+2Y)=0$

so it is a pair of straight lines.

\item[ii)]
$\ xy+2x+y=0 \hspace{0.5in} B^2-4AC=1>0$

so we have a hyperbola or a pair of lines.

$x=X+h \hspace{0.3in} y=Y+k$ gives

$ (X+h)(Y+k)+2(X+h)+(Y+k)=0$

$ XY+X(k+2)+Y(h+1)+hk+2h+k=0$

we choose $h=-1 \hspace{0.3in} k=-2$ so $XY-2=0$.  A rectangular
hyperbola.

Now rotate through $\alpha$

$ X=\xi\cos\alpha-\eta\sin\alpha \hspace{0.5in}
Y=\xi\sin\alpha+\eta\cos\alpha$

$\ds
XY=(\xi\cos\alpha-\eta\sin\alpha)(\xi\sin\alpha+\eta\cos\alpha)$

$\ds
\hspace{0.5in}=\xi^2\frac{1}{2}\sin2\alpha-\eta^2\frac{1}{2}\sin2\alpha+
\xi\eta\cos2\alpha=2$

choose $\alpha$ so that $\cos2\alpha=0$

so $\ds 2\alpha=\frac{\pi}{2} \hspace{0.3in} \alpha=\frac{\pi}{4}$

$XY-2=0$ becomes $\xi^2-\eta^2=4$


DIAGRAM


\end{itemize}

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