Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1

Question

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 latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are

Aicharya · Accepted Answer

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} $

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 latus…

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