$\text{ Let }P\left(x_1,y_1\right)\text{ and }Q\left(x_2,y_2\right),y_1<0,y_2<0\text{ , be the endpoints of the latus rectum of the ellipse }{x}^2+4y^2=4.\text{ The }$

$\mathrm{equations} \mathrm{of} \mathrm{parabolas} \mathrm{with} \mathrm{latus}\mathrm{rectum} PQ\text{ are }$

The given ellipse is

$\begin{array}{l}\frac{x^2}{4}+\frac{y^2}{1}=1\\ b^2=a^2\left(1-e^2\right)\\ \text{or} e=\frac{\sqrt{3}}{2}\end{array}$

Hence, the endpoints P and Q of the latus rectum are given by

$\begin{array}{l}P\equiv \left(\sqrt{3},-\frac{1}{2}\right)\\ \text{and} Q\equiv \left(-\sqrt{3},-\frac{1}{2}\right)\end{array}$

The coordinates of the midpoint of PQ are

$R\equiv \left(0,-\frac{1}{2}\right)$

$\text{ Length of latus rectum, }PQ=2\sqrt{3}$

$\text{ Hence, two parabolas are possible whose vertices are }$

$S\left(0,\frac{\sqrt{3}}{2}-\frac{1}{2}\right)\text{ and }T\left(0,-\frac{\sqrt{3}}{2}-\frac{1}{2}\right)$

The equation of the parabolas are

$\begin{array}{l}{x}^2=2\sqrt{3}\left(y+\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\\ \text{and }{x}^2=-2\sqrt{3}\left(y-\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\\ \text{or } x^2-2\sqrt{3}y=3+\sqrt{3}\\ \text{or} x^2+2\sqrt{3}y=3-\sqrt{3}\end{array}$