IIT-JEE 2008 · Paper 1, Mathematics question 8

Ellipse $\frac{x^2}4+y^2=1$: $e=\frac{\sqrt3}2$, foci $(\pm\sqrt3,0)$, semi-latus rectum $\frac12$. The latus-rectum ends with $y<0$ are $P(\sqrt3,-\frac12)$ and $Q(-\sqrt3,-\frac12)$.

Parabolas with latus rectum PQ. Focus at the midpoint $(0,-\frac12)$, latus rectum $4a=2\sqrt3$, so $a=\frac{\sqrt3}2$; the vertex is $\frac{\sqrt3}2$ above or below the focus.

• Opening up, vertex $\left(0,-\frac12-\frac{\sqrt3}2\right)$: $x^2=2\sqrt3\left(y+\frac12+\frac{\sqrt3}2\right)$, i.e. $x^2-2\sqrt3y=3+\sqrt3$.
• Opening down, vertex $\left(0,-\frac12+\frac{\sqrt3}2\right)$: $x^2=-2\sqrt3\left(y+\frac12-\frac{\sqrt3}2\right)$, i.e. $x^2+2\sqrt3y=3-\sqrt3$.

(B), (C)