JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q9

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MathematicsSingle correct+4 / -1Question 9 of 75

Let a focus of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be $S(4,0)$ and its eccentricity be $\dfrac45$ . If the point $P(3,\alpha)$ lies on E and O is the origin, then the area of $\Delta POS$ is equal to:

12/5

14/5

24/5

48/5

Answer and worked solution

Official answer: C

Ellipse. $ae=4$ with $e=\tfrac45$ , so $a=5$ and $b^2=25-16=9$ .

Point P. $\dfrac9{25}+\dfrac{\alpha^2}9=1$ gives $|\alpha|=\dfrac{12}5$ .

Area. $\tfrac12\cdot OS\cdot|\alpha|=\tfrac12\cdot4\cdot\tfrac{12}5=\dfrac{24}5$ .

(C)

Topics:Ellipse

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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 9