JEE Main 2026 — Mathematics Coordinate Geometry

Given the focus of the ellipse $S(4, 0)$S(4,0), we have $ae = 4$ae=4.

Since the eccentricity $e = \dfrac{4}{5}$e=54​, we get $a \left(\dfrac{4}{5}\right) = 4 \Rightarrow a = 5$a(54​)=4⇒a=5.

Using the relation $b^2 = a^2(1 - e^2)$b2=a2(1−e2), we find:

$$b^2 = 25 \left(1 - \dfrac{16}{25}\right) = 9$$b2=25(1−2516​)=9

The equation of the ellipse is $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$25x2​+9y2​=1.

Since the point $P(3, \alpha)$P(3,α) lies on the ellipse, substituting $x = 3$x=3 gives:

$$\dfrac{9}{25} + \dfrac{\alpha^2}{9} = 1$$259​+9α2​=1

$$\dfrac{\alpha^2}{9} = 1 - \dfrac{9}{25} = \dfrac{16}{25}$$9α2​=1−259​=2516​

$$\alpha^2 = \dfrac{144}{25} \Rightarrow |\alpha| = \dfrac{12}{5}$$α2=25144​⇒∣α∣=512​

The coordinates of the vertices of $\triangle POS$△POS are $O(0, 0)$O(0,0), $S(4, 0)$S(4,0), and $P\left(3, \pm \dfrac{12}{5}\right)$P(3,±512​).

The area of $\triangle POS$△POS is $\dfrac{1}{2} \times \text{base} \times \text{height}$21​×base×height.

Taking $OS$OS as the base, the length is $4$4, and the height is the absolute value of the y-coordinate of $P$P, which is $\dfrac{12}{5}$512​.

Area $= \dfrac{1}{2} \times 4 \times \dfrac{12}{5} = \dfrac{24}{5}$=21​×4×512​=524​

Answer: $24/5$24/5