JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
A line passing through the point \(\mathrm{P}(\sqrt{5}, \sqrt{5})\) intersects the ellipse \(\frac{\mathrm{x}^2}{36}+\frac{\mathrm{y}^2}{25}=1\) at \(A\) and \(B\) such that \((P A) .(P B)\) is maximum. Then \(5\left(P A^2+P B^2\right)\) is equal to :
- A \(218\)
- B \(377\)
- C \(290\)
- D \(338\)
Answer & Solution
Correct Answer
(D) \(338\)
Step-by-step Solution
Detailed explanation
Given ellipse is \(\frac{x^2}{36}+\frac{y^2}{25}=1\) Any point on line \(A B\) can be assumed as \(\mathrm{Q}(\sqrt{5}+\mathrm{r} \cos \theta, \sqrt{5}+\mathrm{r} \sin \theta)\) Putting this in equation of ellipse, we get…
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