Let S and S prime 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 prime P ) 2 - SP bullet S prime P =37 , then α 2 +β 2 is equal to :

Mathematics · Ellipse

JEE Main 2026 — 22 January, Evening Shift — Question 16

Let S and $\mathrm{S}^{\prime}$ be the foci of the ellipse $\frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$ 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 :

Option A:
$15$
Option B:
$11$
Option C:
$17$
Option D:
$13$
Correct

Answer: D

Step-by-step solution

\therefore PS + PS' = 2a \Rightarrow PS + PS' = 10

\therefore (PS)^2 + (PS')^2 - PS \cdot PS' = 37

(PS + PS')^2 - 3PS \cdot PS' = 37

100 - 3PS \cdot PS' = 37

3PS \cdot PS' = 63 \Rightarrow PS \cdot PS' = 21

\therefore PS \text{ and } PS' \text{ are } \left(5 \pm \frac{4}{5}\alpha\right)

\therefore PS \cdot PS' = 25 - \frac{16}{25}\alpha^2 = 21

\frac{16}{25}\alpha^2 = 4

\alpha = \frac{5}{2} \Rightarrow \alpha^2 = \frac{25}{4}

\therefore \beta^2 = \frac{27}{4}

\therefore \alpha^2 + \beta^2 = \frac{52}{4} = 13

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Special properties of ellipse