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Ellipse

Section: Mathematics

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Question : 23 of 117

Marks: +1, -0

A line passing through the point

P(\sqrt{5}, \sqrt{5})

intersects the ellipse

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

A

B

(PA) \cdot (PB)

is maximum. Then

5 (PA^2+PB^2)

[3 Apr 2025 Shift 1]

377
218
338
290

Solution:

\;\;\frac{x-\sqrt{5}}{\cos \theta}=\;\frac{y-\sqrt{5}}{\sin \;\theta}=r

\;x = r \cos \theta + \sqrt{5}

\;y = r \sin \;\theta + \sqrt{5}

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

r^2 \left[ 25 \cos^2 \theta + 36 \sin^2 \;\theta \right] + r \left[ 50 \sqrt{5} \cos \theta + 72 \sqrt{5} \sin \;\theta \right]

+25[5]+36[5]=900

\;r_1 r_2 = \frac{900-305}{25+11 \sin^2 \theta}, (r_1 r_2)_{\max} = \frac{595}{25} = \frac{119}{5}

\; \left| r_1 \cdot r_2 \right|_{\max} = \left( \frac{595}{25} \right) = \left( \frac{119}{5} \right) \;\text{ at }\; \theta = 0

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

\;\text{ 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}}

\;PA^2+PB^2 = (PA+PB)^2 - 2 PA \cdot PB = \left(\frac{24}{\sqrt{5}}\right)^2 - 2 \cdot \frac{119}{5}

\;5 (PA^2+PB^2) = 5 \left( \frac{24^2}{5} - \frac{2.119}{5} \right) = 24^2 - 238

\;=338

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