Let P be a variable point on ellipse x^2 + 3y^2 = 3 , then maximum perpendicular distance of P from

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10-2. Parabola, Ellipse, Hyperbola

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Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$ , then the maximum perpendicular distance of $P$ from the line $x -y = 10$ is

A

$3\sqrt 2 $

B

$4\sqrt 2 $

C

$6\sqrt 2 $

D

$5\sqrt 2 $

Solution

$\frac{x^{2}}{3}+\frac{y^{2}}{1}=1$

$\frac{d y}{d x}=\frac{-x}{3 y}$

$=\frac{-\sqrt{3} \cos \theta}{3 \sin \theta}$

$=\frac{-\cot \theta}{\sqrt{3}}$

$\frac{-\cot \theta}{\sqrt{3}}=1 \Rightarrow \cot \theta=-\sqrt{3}$

$\Rightarrow \theta=-30^{\circ}$ or $150^{\circ}$

For $Q, \theta=150^{\circ} \Rightarrow Q\left(\frac{-3}{2}, \frac{1}{2}\right)$

Maximum distance $=\left|\frac{\frac{-3}{2}-\frac{1}{2}-10}{\sqrt{2}}\right|=6 \sqrt{2}$

Standard 11

Mathematics

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