\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

By transforming co-ordinates reduce the following conics to
standard form.  Sketch them with the original x-y axis
\begin{description}
\item[(i)] $\ds 194x^2 + 120xy + 313y^2 + 776x + 240y + 607 = 0$
\item[(ii)] $\ds 9x^2 + 6xy + y^2 = 4$
\item[(iii)] $\ds 54x^2 - 144xy + 96y^2 + 15y +20x = 0$
\item[(iv)] $\ds x^2 + y^2 - 2x + 4y +9 = 0$


\vspace{.25in}

{\bf Answer}

\item[(i)] $$ 194x^2 + 120xy + 313y^2 + 776x + 240y + 607 = 0$$

${}$

$\ds B^2 - 4AC = 120^2 - 4 \cdot 194 \cdot 194 \cdot 313 <0$
Elliptical type.

Let $x = \xi + h$ and $y = \eta +k$

The equation becomes

$194(\xi + h)^2 + 120(\xi + h)(\eta +k) + 313(\eta +k)^2 + 776(\xi
+ h) + 240(\eta +k) + 607 = 0$

Choose $h$ and $k$ so that the linear terms in $\xi$ and $\eta$
vanish.

$\begin{array}{c} \xi: \hspace{.5in} 338h +120k+776=0 \\ \eta:
\hspace{.5in} 120h+626k+240=0 \end{array} \Rightarrow k = 0, \,
h=-2$

The equation becomes $$194 \xi^2 + 120 \xi \eta + 313 \eta^2 - 169
= 0$$

Now rotate the axis through an angle alpha.  \begin{eqnarray*} \xi
& = & X \cos \alpha - Y\sin \alpha \\ \eta & = & X \sin \alpha + Y
\cos \alpha \end{eqnarray*}

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\put(0,1){\line(1,1){2}}

\put(2,1){\line(0,1){2}}

\put(0.6,1.2){\makebox(0,0){$2\alpha$}}

\put(1,0.6){\makebox(0,0){119}}

\put(1,2.4){\makebox(0,0){169}}

\put(2.5,2){\makebox(0,0){-120}}

\put(1.8,1){\line(0,1){.2}}

\put(1.8,1.2){\line(1,0){.2}}

\end{picture}


Then the equation becomes:

\begin{eqnarray*} X^2(194\cos^2\alpha + 120 \cos \alpha \sin
\alpha + 313 \sin^2 \alpha ) \\ + Y^2(194\cos^2\alpha - 120 \cos
\alpha \sin \alpha + 313 \sin^2 \alpha ) \\ + XY(120(\cos^2\alpha
- \sin^2\alpha) + 238 \sin \alpha \cos \alpha) - 169  & = & 0
\end{eqnarray*}

Choose $\alpha$ so that the coefficient of XY =0

i.e. $\ds 199 \sin 2\alpha = -120 \cos 2 \alpha \Rightarrow \sin
2\alpha = -\frac{120}{169} \, \cos 2\alpha = \frac{119}{169}$

The the equation becomes $$X^2 + 2Y^2 = 1$$


PICTURE

${}$

${}$

\item[(ii)] $$ 9x^2 + 6xy + y^2 = 4$$

${}$

$\ds B^2 - 4AC = 0$ Therefore we have a parabolic Type.

So rotate through angle $\alpha.$

\begin{eqnarray*} x & = & X \cos \alpha - Y \sin \alpha \\ y & = &
X \sin \alpha + Y \cos \alpha \end{eqnarray*}

The equation becomes \begin{eqnarray*} X^2(9\cos^2 \alpha + 6\cos
\alpha \sin \alpha + \sin^2\alpha) \\ + Y^2(9\sin^2 \alpha -
6\sin\alpha \cos\alpha + \cos^2\alpha) \\ +XY(6\cos 2\alpha -
8\sin2\alpha) & = & 4\end{eqnarray*}

Choose $\alpha$ such that $\ds \sin 2\alpha = \frac{6}{10}, \,
\cos 2\alpha= \frac{8}{10} $

\begin{eqnarray*} {\rm So \ \ } \cos^2 \alpha & = &
\frac{1}{2}(1+\cos2\alpha) = \frac{9}{10} \\ \sin^2 \alpha & = &
\frac{1}{2} (1-\cos2\alpha) = \frac{1}{10} \end{eqnarray*}

Then the equation becomes $$5X^2 = 2 {\rm \ \ Parallel\ lines.}$$



${}$

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\put(0,3){\vector(1,0){6}}

\put(3,6.3){\makebox(0,0){y}}

\put(6.3,3){\makebox(0,0){x}}

\put(0,2){\vector(3,1){6}}

\put(4,0){\vector(-1,3){2}}

\put(2,6.3){\makebox(0,0){Y}}

\put(6.3,4){\makebox(0,0){X}}

\put(4.5,0){\line(-1,3){1.9}}

\put(3.5,0){\line(-1,3){1.8}}

\put(1.5,4){\makebox(0,0){y=-3x+2}}

\put(3.9,3.8){\makebox(0,0){y=-3x-2}}



\end{picture}

${}$

\item[(iii)] $$ 54x^2 - 144xy + 96y^2 + 15y +20x = 0$$

${}$

$\ds B^2 - 4AC =0$ Parabolic type.

\begin{eqnarray*} x & = & X \cos \alpha - Y \sin \alpha \\ y & = &
X \sin \alpha + Y \cos \alpha \end{eqnarray*}

We find that we need $\ds 7\sin 2\alpha = 24\cos 2\alpha$ $$sin
2\alpha = \frac{24}{25}, \, \, \, \cos2\alpha = \frac{7}{25}$$ So
$\ds \cos^2\alpha = \frac{1}{2}(1+\cos 2\alpha ) = \frac{16}{25}
\hspace{.2in} \cos \alpha = \frac{4}{5} , \, \, \, \sin \alpha =
\frac{3}{5}$

The equation becomes $$X + 6Y^2 = 0 \Rightarrow y^2 =
-\frac{X}{6}$$


${}$

PICTURE

${}$
\item[(iv)] \begin{eqnarray*} x^2 + y^2 - 2x + 4y +9 & = & 0 \\
(x-1)^2 + (y+2)^2 + 4 & = & 0 \\ {\rm No\ real\ locus}
\end{eqnarray*}






\end{description}
\end{document}
