Let a focus of the ellipse E: x²/a² + y²/b² = 1 be S(4,0) and its eccentricity be 4/5 . If the point P(3,α) lies on E and O is the origin, then the area of ΔPOS is equal to:

Mathematics · Ellipse

JEE Main 2026 — 5 April, Morning Shift — Question 32

Let a focus of the ellipse E: $x²/a² + y²/b² = 1$ 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 $ΔPOS$ is equal to:

Option A:
12/5
Option B:
14/5
Option C:
24/5
Correct
Option D:
48/5

Answer: C

Step-by-step solution

$\mathrm{a}>\mathrm{b}$ Focus $(\mathrm{ae}, 0)=(4,0)$ $\therefore \mathrm{ae}=4$ $\mathrm{a}\left(\frac{4}{5}\right)=4 \Rightarrow \mathrm{a}=5$ $\because b^{2}=a^{2}\left(1-e^{2}\right)$ $b^{2}=25\left(1-\frac{16}{25}\right)$ $\therefore \mathrm{E}: \frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$ ∵ P lies on the ellipse E $\therefore \frac{9}{25}+\frac{\alpha^{2}}{9}=1$ $\therefore \alpha= \pm \frac{12}{5}$ ∴ Area of $\Delta_{\mathrm{POS}}=\frac{1}{2}(\mathrm{OS})(\mathrm{PN})$ $=\frac{1}{2} \times 4 \times \frac{12}{5}=\frac{24}{5}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse