On the ellipse frac{{{x^2}}}{{18}} + frac{{{y | vedclass

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Question

Let $\mathrm{M}$ be $(\sqrt{18} \cos 	heta, \sqrt{8} \sin 	heta) .$ As shortest

distanoe lies along common normal

Given line will be parallel

Vedclass · Accepted Answer

$(-3,2)$

Vedclass · Answer

Let $\mathrm{M}$ be $(\sqrt{18} \cos 	heta, \sqrt{8} \sin 	heta) .$ As shortest

distanoe lies along common normal

Given line will be parallel to the tangent at $\mathrm{M}$

$\frac{x \cos 	heta}{\sqrt{18}}+\frac{y \sin 	heta}{\sqrt{8}}=1$ is parallel to

$2 x-3 y+25=0$

$\Rightarrow-\frac{\sqrt{8} \cos 	heta}{\sqrt{18} \sin 	heta}=\frac{2}{3} $

$\Rightarrow 	an 	heta=-1$

$\Rightarrow 	heta=\frac{3 \pi}{4} \Rightarrow$ point $M$ is $(-3,2)$

On the ellipse $\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{8} = 1$ the point $M$ nearest to the line $2x - 3y + 25 = 0$ is

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