JEE Main 2024 — Ellipse Question with Solution

Let $Z$ be the foot of perpendicular from vertex to directrix of parabola,

Now, finding $Z$ we get,

$\Rightarrow \frac{x-2}{2}=\frac{y-3}{1}=-\frac{\left(4+3\right)-6}{4+1}$

$\Rightarrow \frac{x-2}{2}=\frac{y-3}{1}=\frac{-1}{5}$

$\Rightarrow \frac{x-2}{2}=\frac{-1}{5}, \frac{y-3}{1}=\frac{-1}{5}$

$\Rightarrow x=\frac{12}{5}, y=\frac{16}{5}$

$\Rightarrow Z\equiv \left(\frac{12}{5},\frac{16}{5}\right)$

Eccentricity of ellipse is given as $\frac{1}{\sqrt{2}}$.

Now, finding $b^2$ using eccentricity formula we get,

$\Rightarrow b^2=a^2\left(1-e^2\right)=\frac{a^2}{2}$

So, equation of ellipse will be,

$\Rightarrow \frac{144}{25a^2}+\frac{256}{25\times {\displaystyle \frac{a^2}{2}}}=1$ {as $\left(x,y\right)\equiv \left(\frac{12}{5},\frac{16}{5}\right)$ }

$\Rightarrow \frac{144}{25a^2}+\frac{512}{25a^2}=1$

$\Rightarrow \frac{656}{25a^2}=1$

$\Rightarrow a^2=\frac{656}{25}$

$\Rightarrow b^2=\frac{328}{25}$

Length of latus rectum is given by,
$L=\frac{2b^2}{a}$.

$\Rightarrow L=\frac{2\times {\displaystyle \frac{328}{25}}}{{\displaystyle \frac{\sqrt{656}}{5}}}$

$\Rightarrow L=\frac{\sqrt{656}}{5}$

$\Rightarrow L^2=\frac{656}{25}$