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 $x^2+4y^2=4$. 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 $Q\left(-\sqrt{3},-\frac{1}{2}\right)($
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 $x^2+2\sqrt{3}y=3-\sqrt{3}$.