Question

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 $PQ$ 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. $x^{2}-2 \sqrt{3} y=3-\sqrt{3}$

Answer 1

Answer: (B), (C)

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$
$b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow e=\frac{\sqrt{3}}{2}$
$\Rightarrow P \left(\sqrt{3},-\frac{1}{2}\right)$ and $Q \left(-\sqrt{3},-\frac{1}{2}\right)\left(\right.$
given $y _{1}$ and $y _{2}$ less than $0)$
Co-ordinates of mid-point of $PQ$ are
$R \equiv\left(0,-\frac{1}{2}\right) $
$PQ =2 \sqrt{3}=$ 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 $x^{2}+2 \sqrt{3} y=3-\sqrt{3}$.