Let P (x_1, y_1) and Q(x_2, y_2), y_1 < 0, y_2 < 0 , be the end...

The equation $x^2+4 y^2=4$ represents an ellipse with 2 and 1 as semi-major and semi-minor axes and eccentricity $\frac{\sqrt{3}}{2}$. Thus, the ends of latusrect are \left(\sqrt{3}, \frac{1}{2}\right)$and \left(\sqrt{3},-\frac{1}{2}\right), \left(-\sqrt{3},-\frac{1}{2}\right) and \left(\sqrt{3},-\frac{1}{2}\right). <img src="/api/proxy-image?url=httpsAccording to the question, we consider only$P \left(-\sqrt{3},-\frac{1}{2}\right)$and$Q \left(\sqrt{3},-\frac{1}{2}\right). Now, $\quad P Q=2$\sqrt{3} Thus, the coordinates of the vertex of the parabolas are $A \left(0, \frac{-1+\sqrt{3}}{2}\right)$ and A^{\prime} \left(0, \frac{-1-\sqrt{3}}{2}\right) and corresponding equations are and (x-0)^2=-4 \cdot \frac{\sqrt{3}}{2}\left(y+\frac{1-\sqrt{3}}{2}\right) i.e.,(x-0)^2=4 $\cdot \frac{\sqrt{3}}{2} \lleft(y-\frac{-1-\sqrt{3}}{2})$and $$\begin{aligned}& x²+2 \ssqrt{3}y=3-\sqrt{3}\\ & x^2-2 \ssqrt{3}y=3+\sqrt{3} \end{aligned}$$