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 $\mathrm{P}(3, \alpha)$ lies on E and O is the origin, then the area of $\triangle \mathrm{POS}$ is equal to :

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Ellipse

Let a focus of the ellipse

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

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

△ POS

is equal to :

Answer: (C)

From

a e = 4

and

e = \frac{4}{5}

we get

a = 5

, and

b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9

, so the ellipse is

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

. Since

P (3, α)

lies on it,

\frac{9}{25} + \frac{α ^{2}}{9} = 1

, giving

α^{2} = \frac{144}{25}

and

∣ α ∣ = \frac{12}{5}

. Taking

O S = 4

as the base along the

x

-axis and

∣ α ∣

as the height, the area of

△ P O S

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

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option