Mathematics · Ellipse

JEE Main 2025 — 2 April, Morning Shift — Question 26

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{9}=1$ and $P$ be a point on the ellipse, then $\min \left(S P \cdot S^{\prime} P\right)+$ $\max \left(S P \cdot S^{\prime} P\right)$ is equal to

Option A:
$27$
Correct
Option B:
$3(1+\sqrt{2})$
Option C:
$3(6+\sqrt{2})$
Option D:
$9$

Answer: A

Step-by-step solution

$a=3 \sqrt{2}, b=3$

\Rightarrow \quad e=\frac{1}{\sqrt{2}}

$P S \cdot P S^{\prime}=2 a=6 \sqrt{2}$

$\frac{P S+P S^{\prime}}{2} \geq \sqrt{P S \cdot P S^{\prime}}$

$\Rightarrow\left(P S \times P S^{\prime}\right) \max =18$

Minima happens when $P$ lies on major axis

$\Rightarrow \quad P=(3 \sqrt{2}, 0)$

$P S=(3 \sqrt{2}-3) \cdot P S^{\prime}=(3 \sqrt{2}+3)$

( $P S . P S^{\prime}$ ) $\min =9$

$({PS.PS}')min + (PS•PS') max$ = 27

Option (1)

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Special properties of ellipse