Let the point L lying in the first quadrant be one end of a latus rectum of the ellipse…
Let the point L lying in the first quadrant be one end of a latus rectum of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P$ and $Q$ be the points where the normal drawn at $\mathrm{L}$ to this given ellipse meets the major axis and the minor axis. Then the distance between $\mathrm{P}$ and $\mathrm{Q}$ is
$\frac{\sqrt{5}}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2 \sqrt{2}}$
$\frac{\sqrt{5}}{2}$
Solution
Given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
Here, $a=2, b=\sqrt{3}$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
$\because \mathrm{L}$ is lying in the first quadrant
$\therefore \mathrm{L}=\left(a e, \frac{b^2}{a}\right)=\left(2 \times \frac{1}{2}, \frac{3}{2}\right)=\left(1, \frac{3}{2}\right)$
Now, $y^{\prime}=-\frac{b^2}{a^2}, \frac{x}{y}=-\frac{3}{4} \times \frac{1}{\frac{3}{2}}=-\frac{1}{2}$
$m=-\frac{1}{y^{\prime}}=2$
Equation of normal at $\mathrm{L}$ is
$\begin{aligned} & \left(y-\frac{3}{2}\right)=2(x-1) \Rightarrow 2 x-y-\frac{1}{2}=0 \\ & \Rightarrow \frac{x}{\frac{1}{4}}+\frac{y}{\left(-\frac{1}{2}\right)}=1\end{aligned}$
Which meet to the minor axis at $\left(0,-\frac{1}{2}\right)$ and major axis at $\left(\frac{1}{4}, 0\right)$.
$\therefore \mathrm{P} \equiv\left(\frac{1}{4}, 0\right) \& \mathrm{Q} \equiv\left(0,-\frac{1}{2}\right)$
Distance between $P$ \& $Q$ is
$D=\sqrt{\left(\frac{1}{4}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{1}{16}+\frac{1}{4}}=\frac{\sqrt{5}}{4}$