JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

Let a focus of the ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be $S (4, 0)$ and its eccentricity be $\frac{4}{5}$ . If the point $P (3, α)$ lies on $E$ and $O$ is the origin, then the area of $△ P O S$ is equal to :

A
$\frac{12}{5}$
B
$\frac{14}{5}$
C
$\frac{24}{5}$
D
$\frac{48}{5}$

Solution

Since the focus lies on the $x$ -axis at $(4, 0)$ , $a e = 4$ . With $e = \frac{4}{5}$ , we get $a = 5$ and $b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$ .

The ellipse is $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . Substituting $x = 3$ : $\frac{9}{25} + \frac{α ^{2}}{9} = 1$ , giving $α^{2} = \frac{144}{25}$ and $∣ α ∣ = \frac{12}{5}$ .

Triangle $P O S$ has vertices $O (0, 0)$ , $S (4, 0)$ , $P (3, α)$ . With $O S$ as base $(= 4)$ and height $∣ α ∣$ :

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

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