JEE Main 2026 — Ellipse Question with Solution

Given the focus of the ellipse $S(4,0)$, we have $ae=4$.

Since the eccentricity $e=\frac{4}{5}$, we get $a\left(\frac{4}{5}\right)=4\Rightarrow a=5$.

Using the relation $b^2=a^2(1-e^2)$, we find:

$b^2=25\left(1-\frac{16}{25}\right)=9$

The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.

Since the point $P(3,\alpha )$ lies on the ellipse, substituting $x=3$ gives:

$\frac{9}{25}+\frac{{\alpha }^2}{9}=1$

$\frac{{\alpha }^2}{9}=1-\frac{9}{25}=\frac{16}{25}$

${\alpha }^2=\frac{144}{25}\Rightarrow |\alpha |=\frac{12}{5}$

The coordinates of the vertices of $△POS$ are $O(0,0)$, $S(4,0)$, and $P\left(3,\pm \frac{12}{5}\right)$.

The area of $△POS$ is $\frac{1}{2}\times \text{base}\times \text{height}$.

Taking $OS$ as the base, the length is $4$, and the height is the absolute value of the y-coordinate of $P$, which is $\frac{12}{5}$.

Area $=\frac{1}{2}\times 4\times \frac{12}{5}=\frac{24}{5}$

Answer: $24/5$