Ellipse

We have $x^2+4y^2$ = 4 $\frac{x^2}{4}+\frac{y^2}{1}$ = 1 and $b^2$ = $a^2(1-e^2)$ 1 = $(4(1-e^2)$ ⇒ e = $\frac{\sqrt{3}}{2}$ Therefore, the points P and Q are $P\left(\sqrt{3},-\frac{1}{2}\right)$ ; Q $\left(-\sqrt{3} , -\frac{1}{2}\right)$ Since P and Q are the endpoints of latus rectum of parabola The midpoint of PQ is the focus of parabola. Hence, focus = $\left(0,-\frac{1}{2}\right)$ Hence, the length of latus rectum = 4a. Therefore, 4a = $2\sqrt{3}$ ⇒ a = $\frac{\sqrt{3}}{2}$ Hence, the two parabolas 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 equation of the parabolas are given as follows: $x^2$ = $2\sqrt{3} \left[ y-\left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) \right]$ and $x^2$ = $2\sqrt{3}\left[ y-\left( -\frac{\sqrt{3}}{2} - \frac{1}{2} \right) \right]$