Mathematics - Ellipse Question with Solution | TestHub

We have,

$x^2+3y^2=6$

$\Rightarrow \frac{x^2}{6}+\frac{y^2}{2}=1$

Centre of an ellipse is $\left(0,0\right)$.

Let the foot of perpendicular be $(h,k)$.

Equation of tangent in slope form to the standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$y=mx\pm \sqrt{a^2{m}^2+b^2}$.

Hence, equation of tangent with slope $m$ passing $(h,k)$ is 

$y=mx\pm \sqrt{6m^2+2}$

$\Rightarrow k=mh\pm \sqrt{6m^2+2}$, where $m=-\frac{h}{k}$.

$\Rightarrow k=h\left(-\frac{h}{k}\right)\pm \sqrt{6m^2+2}$

$\Rightarrow \sqrt{\frac{6h^2}{k^2}+2}=\frac{h^2+k^2}{k}$

$\Rightarrow 6h^2+2k^2={\left(h^2+k^2\right)}^2$

So, required locus is

$6x^2+2y^2={\left(x^2+y^2\right)}^2$.