Answer:
Option D
Explanation:
Equation of ellipse is $x^{2}+3y^{2}=6 $ or $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$
Equation of the tangent is
$\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$
Let (h,k) be any point on the locus.
$\therefore$ $\frac{h}{a}\cos\theta+\frac{k}{b}\sin\theta=1$ .....(i)
Slope of the tangent $\frac{-b}{a}\cot\theta$
Slope of perpendicular drawn from centre (0,0) to (h,k) is k/h
Since, both the lines are perpendicular
$\therefore$ $\left(\frac{k}{h}\right)\times(\frac{-b}{a}\cot\theta)=-1$
$\Rightarrow$ $\frac{\cos\theta}{ha}=\frac{\sin\theta}{kb}=\alpha$ [say]
$\Rightarrow$ $\cos\theta=\alpha ha$
$\sin \theta=\alpha kb$
From Eq(i) , we get
$\frac{h}{\alpha}(\alpha ha+\frac{k}{b}(\alpha kb)=1$
$\Rightarrow$ $h^{2}\alpha+k^{2}\alpha=1$
$\Rightarrow$ $\alpha=\frac{1}{h^{2}+k^{2}}$
Also, $\sin^{2}\theta+\cos^{2}\theta=1$
$\Rightarrow$ $(\alpha kb)^{2}+(\alpha ha)^{2}=1$
$\Rightarrow$ $\alpha^{2}k^{2}h^{2}+\alpha^{2}h^{2}a^{2}=1$
$\Rightarrow$ $\frac{k^{2}b^{2}}{(h^{2}+k^{2})^{2}}+\frac{h^{2}a^{2}}{(h^{2}+k^{2})^{2}}=1$
$\Rightarrow$ $\frac{2k^{2}}{(h^{2}+k^{2})^{2}}+\frac{6h^{2}}{(h^{2}+k^{2})^{2}}=1$
[$\therefore$ a2=6, b2=2]
$\Rightarrow$ $6x^{2}+2y^{2}=(x^{2}+y^{2})^{2}$
[Replaceing k by y and h by x]
