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{\bf Question}

\begin{description}
\item[(i)] Find the parabola of the form $Ax +By^2+cy +D=0$
through the three points $P_1(-1,0), \, P_2(2,1), \, P_3(1,-1).$
Find the vertex and the axis of the parabola and sketch.
\item[(ii)] Find a point on intersection of the parabolas $$ r =
\frac{1}{1-\cos \theta} \hspace{.2in} r = \frac{3}{1+\cos
\theta}$$ and find the angle between the tangents to these curves
at this point.
\end{description}

\vspace{.25in}

{\bf Answer}
\begin{description}
\item[(i)]
\begin{eqnarray*} Ax + By^2 +Cy +D & = & 0 \\ {\rm So}\hspace{1in} A + D & =
& 0 \\ 2A + B + C + D & = & 0 \\ A + B -C + D & = & 0
\end{eqnarray*}
Choose $A=D=2$ then $B+C=-6,\,$ $B-C=-4$, so $B = -5, \,$ $C=-1$

So the parabola is $$2x - 5y^2 - y +2=0$$ or $$\left(y +
\frac{1}{10} \right)^2 = \frac{2}{5} \left(
x+\frac{41}{40}\right)$$

${}$

\begin{center}
\epsfig{file=g79-1-1.eps, width=45mm}
\end{center}

${}$

Vertex $ = \left( -\frac{41}{40} , -\frac{1}{10} \right)$ axis
$y=-\frac{1}{10}$

${}$

\item[(ii)] The parabolas intersect where $\ds
\frac{1}{1-\cos\theta} = \frac{3}{1+\cos \theta}$

giving $\ds \cos \theta = \frac{1}{2} \hspace{.2in} \theta = \pm
\frac{\pi}{3}.$

When $\cos \theta = \frac{1}{2} , \, r = 2$

So points of intersection are $\ds \left(2, \pm
\frac{\pi}{3}\right)$

\begin{center}
\epsfig{file=g79-1-2.eps, width=50mm}
\end{center}

For $\ds r = \frac{1}{1-\cos\theta} \hspace{.2in}
\frac{dr}{d\theta} = -\frac{\sin \theta}{(1-\cos \theta)^2}$

$\ds \frac{1}{r}\frac{dr}{d\theta} = -\frac{\sin
\theta}{1-\cos\theta} = -\sqrt3 \hspace{.2in} \tan \phi_2 =
-\frac{1}{\sqrt3} \phi_2 = 150^\circ$

For $\ds r = \frac{3}{1+\cos\theta} \hspace{.2in}
\frac{1}{r}\frac{dr}{d\theta} = \frac{\sin\theta}{1+\cos\theta}
\hspace{.2in} \tan \phi_1 = \sqrt3 \hspace{.2in} \phi_1 =
60^\circ$

So the angle between the tangents is $90^\circ$


\end{description}
\end{document}