If S and are the foci of the ellipse and P be a point on the ellipse, then is equal to : [2025]

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Solution

Q.
If S and $S'$ are the foci of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$ and P be a point on the ellipse, then $\min (S P \cdot S' P) + \max (S P \cdot S' P)$ is equal to : [2025]

1 27

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

3 9

4 $3 (1 + \sqrt{2})$

Ans.
(1)

We have, $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$

$\Rightarrow a = 3 \sqrt{2} and b = 3$

Since, $P S + P S' = 2 a = 2 \times 3 \sqrt{2} = 6 \sqrt{2}$

Also, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow 9 = 18 (1 - e^{2})$

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

Equation of Directrix is $X = \frac{a}{e} = \frac{3 \sqrt{2}}{\frac{1}{\sqrt{2}}} = 6$

$S P \cdot S P' =$ $| \frac{1}{\sqrt{2}} (3 \sqrt{2} \cos θ - 6) \cdot \frac{1}{\sqrt{2}} (3 \sqrt{2} \cos θ + 6) |$

$= \frac{1}{2}$ $| 18 \cos^{2} θ - 36 |$ $=$ $| 9 \cos^{2} θ - 18 |$

${(S P \cdot S P')}_{\max} = 18; {(S P \cdot S P')}_{\min} = 9$ [ $∵ \cos^{2} x \in [0, 1]$ ]

$∴$ Required sum = 27.

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