If S and S^{prime} are foci of ellipse frac{x^2}{18}+frac{y^2}{9}=1 and P be a point on ellipse, the

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10-2. Parabola, Ellipse, Hyperbola

hard

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 $\min \left(S P . S^{\prime} P\right)+$ $\max \left( SP . S ^{\prime} P \right)$ is equal to :

A$3(1+\sqrt{2})$

B$3(6+\sqrt{2})$

C$9$

D$27$

(JEE MAIN-2025)

Solution

image
$PS+PS^{\prime}=2 \times 3 \sqrt{2}$
$b^2=a^2\left(1-e^2\right) \Rightarrow 9=18\left(1-e^2\right)$
$\Rightarrow e=\frac{1}{\sqrt{2}}$
Directrix $x =\frac{ a }{ e }=\frac{3 \sqrt{2}}{\frac{1}{\sqrt{2}}}=6$
$PS \cdot P S^{\prime}=\left|\frac{1}{\sqrt{2}}(3 \sqrt{2} \cos \theta-6) \frac{1}{\sqrt{2}}(3 \sqrt{2} \cos \theta+6)\right|$
$=\frac{1}{2}\left|18 \cos ^2 \theta-36\right|$
$( PS \cdot PS )_{\max }=18 ;( PS \cdot PS )_{\min }=9$
$sum =27$

Standard 11

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