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\begin{document}


{\bf Question}

\begin{description}
\item[(a)] Reduce the conic $$6x^2 - 24xy - y^2 +60x - 20y +
45=0$$ to standard form, and identify its type.
\item[(b)] Find the eccentricity, foci, and semi latus rectum of
the ellipse $$3x^2 + 4y^2 = 1$$ Hence write down all its polar
equation referred to a focus as origin and major axis as initial
line.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $6x^2 - 24xy - y^2 +60x - 20y + 45=0$

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

$6(\xi +h)^2 - 24(\xi +h)(\eta + k) - (\eta + k)^2 +60(\xi +h) -
20(\eta + k) + 45=0$

$6\xi^2 - 24\xi \eta - \eta^2 +\xi(12h - 24k +60) - \eta(24h +2k
+20) +6h^2 - 24hk - k2 + 60h - 29k +45 = 0$

Choose $h,k$ so that

$\left.\begin{array}{c} 12h - 24k +60 \\ 24h +2k +20 \end{array}
\right\} k = 2, h = -1$

So the equation becomes $$6\xi^2 - 24\xi\eta - \eta^2=5$$

${}$

Now $\xi = X \cos \alpha - Y \sin \alpha $ and $\eta = X \sin
\alpha + Y \cos \alpha$

gives \begin{eqnarray*} 5 & = & 6(X^2 \cos^2 \alpha - 2XY \cos
\alpha \sin \alpha  + Y^2 \sin^2 \alpha \\ & & -24(X\cos \alpha -
Y\sin \alpha )(X\sin\alpha + Y\cos \alpha) \\ & & -(X^2 \sin^2
\alpha + 2XY \cos \alpha \sin \alpha + Y^2 \cos ^2 \alpha) \\ {}
\\ 5 & = & X^2(6 \cos ^2\alpha - 24 \cos \alpha \sin \alpha -
\sin^2 \alpha) \\ & & Y^2(6 \sin^2 \alpha + 24 \cos \alpha \sin
\alpha  - \cos^2 \alpha) \\ & & XY(-12 \cos \alpha \sin \alpha
-24(\cos^2\alpha - \sin^2\alpha) -2(\cos \alpha \sin\alpha)
\end{eqnarray*}


We want the coefficient of XY to be zero  $-24 \cos 2\alpha - 7
\sin 2\alpha =0$ So $\tan 2\alpha = -\frac{24}{7}$


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\put(0.9,.2){-7}

\put(.9,1.2){25}

\put(2.2,.75){$24$}

\put(.7,.6){$2\alpha$}

\end{picture}

Choose $\cos 2 \alpha = -\frac{7}{25}$ and $\sin\alpha =
\frac{24}{25}$

$\cos^2\alpha = \frac{1}{2}(\cos 2\alpha +1) = \frac{9}{25}
\hspace{.2in} \sin^2 \alpha = \frac{16}{25}$

$$3Y^2 - 2X^2=1$$

\item[(b)] Eccentricity $\ds = \frac{1}{2}$

Foci $\ds  = \left(\pm \frac{1}{2\sqrt3},0\right)$

$\ds l = \frac{\sqrt 3}{4}$

Polar equation $\ds \frac{\sqrt3}{4r} = 1 \pm \frac{1}{2} \cos
\theta$

\end{description}


\end{document}