Mathematics - Ellipse Question with Solution | TestHub

The equation $x^2+4y^2=4$ 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 $\left(\sqrt{3},-\frac{1}{2}\right)$.
According to the question, we consider only $P\left(-\sqrt{3},-\frac{1}{2}\right)$ and $Q\left(\sqrt{3},-\frac{1}{2}\right)$. Now, $PQ=2\sqrt{3}$ 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., $(x-0{)}^2=4\cdot \frac{\sqrt{3}}{2}\left(y-\frac{-1-\sqrt{3}}{2}\right)$ and $$\begin{array}{rl}& x^2+2\sqrt{3}y=3-\sqrt{3}\\ & x^2-2\sqrt{3}y=3+\sqrt{3}\end{array}$$