A line passing through the point intersects the ellipse at A and B such that (PA)(PB) is maximum. Then is equal to : [2025]

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Solution

Q.
A line passing through the point $P (\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$ at A and B such that (PA) $\cdot$ (PB) is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to : [2025]

1 338

2 377

3 218

4 290

Ans.
(1)

$\frac{x - \sqrt{5}}{\cos θ} = \frac{y - \sqrt{5}}{\sin θ} = r$

$x = r \cos θ + \sqrt{5}$

$y = r \sin θ + \sqrt{5}$

$25 x^{2} + 36 y^{2} = 900, (x, y)$ lie on E

$r^{2} [25 \cos^{2} θ + 36 \sin^{2} θ] + r [50 \sqrt{5} \cos θ + 72 \sqrt{5} \sin θ] + 25 [5] + 36 [5] = 900$

$r_{1} r_{2} = \frac{900 - 305}{25 + 11 \sin^{2} θ}, {(r_{1} r_{2})}_{\max} = \frac{595}{25} = \frac{119}{5}$

${| r_{1} \cdot r_{2} |}_{\max} = (\frac{595}{25}) = (\frac{119}{5}) at θ = 0$

$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$

At $y = \sqrt{5}$

$\frac{x^{2}}{36} + \frac{1}{5} = 1 \Rightarrow \frac{x^{2}}{36} = \frac{4}{5} \Rightarrow x = \pm \frac{12}{\sqrt{5}}$

$P A^{2} + P B^{2} = {(P A + P B)}^{2} - 2 P A \cdot P B = {(\frac{24}{\sqrt{5}})}^{2} - 2 \cdot \frac{119}{5}$

$5 (P A^{2} + P B^{2}) = 5 (\frac{24^{2}}{5} - \frac{2.119}{5}) = 24^{2} - 238 = 338$ .

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