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Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1<0, y_2<0$, be the end points of the latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are
$(A)$ $x^2+2 \sqrt{3} y=3+\sqrt{3}$
$(B)$ $x^2-2 \sqrt{3} y=3+\sqrt{3}$
$(C)$ $x^2+2 \sqrt{3} y=3-\sqrt{3}$
$(D)$ $x^2-2 \sqrt{3} y=3-\sqrt{3}$
$B,D$
$C,A$
$B,C$
$B,C$
Solution
$ \frac{x^2}{4}+\frac{y^2}{1}=1 $
$ b^2=a^2\left(1-e^2\right) $
$ \Rightarrow e=\frac{\sqrt{3}}{2} $
$ \left.\Rightarrow P\left(\sqrt{3},-\frac{1}{2}\right) \text { and } Q\left(-\sqrt{3},-\frac{1}{2}\right) \text { (given } y_1 \text { and } y_2 \text { less than } 0\right)$
Co-ordinates of mid-point of $P Q$ are
$\mathrm{R} \equiv\left(0,-\frac{1}{2}\right) \text {. }$
$P Q=2 \sqrt{3}=$ length of latus rectum.
$\Rightarrow$ two parabola are possible whose vertices are $\left(0,-\frac{\sqrt{3}}{2}-\frac{1}{2}\right)$ and $\left(0, \frac{\sqrt{3}}{2}-\frac{1}{2}\right)$.
Hence the equations of the parabolas are $x^2-2 \sqrt{3} y=3+\sqrt{3}$ and $x^2+2 \sqrt{3} y=3-\sqrt{3}$.
