Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ and $\mathrm{P}(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(\mathrm{SP})^{2}+\left(\mathrm{S}^{\prime} \mathrm{P}\right)^{2}-\mathrm{SP} \bullet \mathrm{S}^{\prime} \mathrm{P}=37$, then $\alpha^{2}+\beta^{2}$ is equal to :

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Ellipse

Let

S

and

S^{'}

be the foci of the ellipse

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

and

P (α, β)

be a point on the ellipse in the first quadrant. If

(SP)^{2} + (S^{'} P)^{2} - SP ∙ S^{'} P = 37

, then

α^{2} + β^{2}

is equal to :

Answer: (D)

For

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

a = 5, b = 3, e = \frac{4}{5}

, and

P S + P S^{'} = 2 a = 10

. Writing

p = P S \cdot P S^{'}

(P S)^{2} + (P S^{'})^{2} - P S \cdot P S^{'} = (P S + P S^{'})^{2} - 3 P S \cdot P S^{'} = 100 - 3 p = 37 \Rightarrow p = 21.

With focal radii

P S = 5 - \frac{4}{5} α, P S^{'} = 5 + \frac{4}{5} α

P S \cdot P S^{'} = 25 - \frac{16}{25} α^{2} = 21

, so

α^{2} = \frac{25}{4}

. From the ellipse,

β^{2} = 9 (1 - \frac{α ^{2}}{25}) = \frac{27}{4}

. Hence

α^{2} + β^{2} = \frac{25}{4} + \frac{27}{4} = 13

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option