Question

Let $P(x_1,y_1)$ and $Q(x_2,y_2);$ where $y_1<0,y_2<0$ be the end points of the latus rectum of the ellipse $x^2+4y^2=4.$ The equation of parabola 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

The correct option is C $x^2+2\sqrt{3}y=3-\sqrt{3}$

$\frac{x^2}{4}+\frac{y^2}{1}=1$

$P=(-ae,-\frac{b^2}{a}),Q=(ae,-\frac{b^2}{a})$

Hence $P=(-\sqrt{3},-\frac{1}{2}),Q=(\sqrt{3},-\frac{1}{2})$

So the focus of parabola is $(0,-\frac{1}{2})$

Also $4a$ is the distance between P and Q

$4a=2\sqrt{3}$

So ${(x-0)}^2=\pm 2\sqrt{3}(y-(\frac{\sqrt{3}}{2}-\frac{1}{2}\left)\right)$

${\left(x\right)}^2=\pm (2\sqrt{3}y-3+\sqrt{3})$

Hence, C is correct.