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]