MCQ
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 $
- ✓$6\sqrt 2 $
- D$5\sqrt 2 $
✓
Answer
Correct option: C.
$6\sqrt 2 $
c
$\frac{x^{2}}{3}+\frac{y^{2}}{1}=1$
$\frac{x^{2}}{3}+\frac{y^{2}}{1}=1$
$\frac{d y}{d x}=\frac{-x}{3 y}$
$=\frac{-\sqrt{3} \cos \theta}{3 \sin \theta}$
$=\frac{-\cot \theta}{\sqrt{3}}$
$\frac{-\cot \theta}{\sqrt{3}}=1 \Rightarrow \cot \theta=-\sqrt{3}$
$\Rightarrow \theta=-30^{\circ}$ or $150^{\circ}$
For $Q, \theta=150^{\circ} \Rightarrow Q\left(\frac{-3}{2}, \frac{1}{2}\right)$
Maximum distance $=\left|\frac{\frac{-3}{2}-\frac{1}{2}-10}{\sqrt{2}}\right|=6 \sqrt{2}$
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