EllipseeasyPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Area of Triangle POS on an Ellipse: 24/5 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a focus of the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 be S(4,0)S(4,0) and its eccentricity be 45\dfrac45. If the point P(3,α)P(3,\alpha) lies on EE and OO is the origin, then the area of △POS\triangle POS is
A125\dfrac{12}{5}
B145\dfrac{14}{5}
C245\dfrac{24}{5}correct
D485\dfrac{48}{5}
Solution
Step 1: ae=4ae=4, e=45⇒a=44/5=5e=\dfrac45\Rightarrow a=\dfrac{4}{4/5}=5. b2=a2(1−e2)b^2=a^2(1-e^2):
b2=25(1−1625)=9 ⇒ E: x225+y29=1.b^2=25\left(1-\frac{16}{25}\right)=9\ \Rightarrow\ E:\ \frac{x^2}{25}+\frac{y^2}{9}=1.
 Step 2: Put P(3,α)P(3,\alpha) in EE: 925+α29=1⇒α29=1625\dfrac{9}{25}+\dfrac{\alpha^2}{9}=1\Rightarrow\dfrac{\alpha^2}{9}=\dfrac{16}{25}.
α2=14425 ⇒ α=±125.\alpha^2=\frac{144}{25}\ \Rightarrow\ \alpha=\pm\frac{12}{5}.
 Step 3: Base OS=4OS=4, height =∣α∣=125=|\alpha|=\dfrac{12}{5}:
Area=12⋅4⋅125=245.\text{Area}=\frac12\cdot4\cdot\frac{12}{5}=\frac{24}{5}.
 Correct answer: (3)
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