JEE Main 2023 — Ellipse Question with Solution

Given,

An ellipse with centre $\left(1, 0\right)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis 

And its minor axis subtends an angle ${60}^{\circ }$ at the foci, 

So, on plotting the diagram we get,

Now from diagram we get,

$\frac{2b^2}{a}=\frac{1}{2}\Rightarrow b^2=\frac{a}{4} ...\left(1\right)$

And $\tan {30}^{\circ }=\frac{b}{ae}$

$\Rightarrow {\left(\frac{1}{\sqrt{3}}\right)}^2={\left(\frac{b}{ae}\right)}^2$

$$$\Rightarrow \frac{1}{3}=\frac{b^2}{a^2-b^2}$

$\Rightarrow a^2-b^2=3b^2$

$\Rightarrow b^2=\frac{a^2}{4} .......\left(2\right)$

Now from equation $\left(1\right) \& \left(2\right)$ we get,

$\Rightarrow a=1, b^2=\frac{1}{4}\Rightarrow b=\frac{1}{2}$

Hence, ${\left(2a+2b\right)}^2=9$