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Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1 < 0, y_2 < 0$, be the end points of the latusrectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latusrectum $P Q$ are
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Verified Answer
The correct answers are:
$x^2-2 \sqrt{3} y=3+\sqrt{3}$
,
$x^2+2 \sqrt{3} y=3-\sqrt{3}$
$x^2-2 \sqrt{3} y=3+\sqrt{3}$
,
$x^2+2 \sqrt{3} y=3-\sqrt{3}$
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 $\left(\sqrt{3},-\frac{1}{2}\right)$.
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, $\quad P Q=2 \sqrt{3}$
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., $(x-0)^2=4 \cdot \frac{\sqrt{3}}{2}\left(y-\frac{-1-\sqrt{3}}{2}\right)$ and
$$
\begin{aligned}
& x^2+2 \sqrt{3} y=3-\sqrt{3} \\
& x^2-2 \sqrt{3} y=3+\sqrt{3}
\end{aligned}
$$
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, $\quad P Q=2 \sqrt{3}$
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., $(x-0)^2=4 \cdot \frac{\sqrt{3}}{2}\left(y-\frac{-1-\sqrt{3}}{2}\right)$ and
$$
\begin{aligned}
& x^2+2 \sqrt{3} y=3-\sqrt{3} \\
& x^2-2 \sqrt{3} y=3+\sqrt{3}
\end{aligned}
$$
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