Question

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ , $y_1<0,y_2<0$, 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)

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

$\frac{x^2}{4}+\frac{y^2}{1}=1$
$b^2=a^2(1-e^2)$
$\Rightarrow e=\frac{\sqrt{3}}{2}$
$\Rightarrow P\left(\sqrt{3},-\frac{1}{2}\right)$ and $Q(-\sqrt{3},-\frac{1}{2})$ (given $y_1$ and $y_2$ less than $0$).
Co-ordinates of mid-point of $PQ$ are
$R\equiv (0,-\frac{1}{2})$
$PQ=2\sqrt{3}=$ length of latus rectum.
$\Rightarrow$ Two parabolas are possible whose vertices are $(0,-\frac{\sqrt{3}}{2}-\frac{1}{2})$ and $(0,\frac{\sqrt{3}}{2}-\frac{1}{2})$
Hence, the equations of the parabolas are $x^2-2\sqrt{3}y=3+\sqrt{3}$ and $x^2+2\sqrt{3}y=3-\sqrt{3}$.