JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 5)
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 :
$3(6+\sqrt{2})$
$3(1+\sqrt{2})$
27
9
Explanation
$$\begin{aligned} & a=3 \sqrt{2}, b=3 \\ & \Rightarrow \quad e=\frac{1}{\sqrt{2}} \end{aligned}$$
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$$\begin{aligned} & P S \cdot P S^{\prime}=2 a=6 \sqrt{2} \\ & \frac{P S+P S^{\prime}}{2} \geq \sqrt{P S \cdot P S^{\prime}} \\ & \Rightarrow\left(P S \times P S^{\prime}\right) \max =18 \end{aligned}$$
Minima happens when $P$ lies on major axis
$$\begin{aligned} & \Rightarrow \quad P=(3 \sqrt{2}, 0) \\ & P S=(3 \sqrt{2}-3) \cdot P S^{\prime}=(3 \sqrt{2}+3) \\ & \left(P S \cdot P S^{\prime}\right) \min =9 \\ & \left(P S \cdot P S^{\prime}\right) \min +\left(P S \cdot P S^{\prime}\right) \max =27 \end{aligned}$$
Option (1)
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