the locus of the foot of perpendicular drawn from the centre of the ellipseX²+ 3y²=6 on any tangent to it is

The given ellipse is:-

$$\begin{gathered} x^2+3y^2=6 \\ \Rightarrow \frac{x^2}{6}+\frac{y^2}{2}=1 \end{gathered}$$

Now, we know that any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by:-

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

So, the equation of the tangent to the given ellipse is:-

$y=mx+\sqrt{6m^2+2}$   -------------------(1)

Now, equation of the line through (0, 0) and perpendicular to (1) is:-

$y-0=\left(-\frac{1}{m}\right)x-0$

$\Rightarrow m=-\frac{x}{y}$   ------------------(2)

Using (2) in (1), we have:-

$y=\left(-\frac{x}{y}\right)x+\sqrt{6\left(\frac{x^2}{y^2}\right)+2}$

$\Rightarrow y^2=-x^2+\sqrt{6x^2+2y^2}$

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

This is the required locus.