Question
What is the Locus of the foot of the perpendicular drawn from the centre of the ellipse $x^2$ +$3y^2$ = 6
I proceeded by assuming a pt $(h,k)$ as the foot of the perpendicular to the tangent drawn and found the slope of the tangent through differentiating the equation of ellipse at that point I then put the values in the point slope form but then realised that I could not eliminate the parameter that is why I am confused on how to proceed thanks in advance please answer asap and tell me if I am doing some trivial mistake .
Th equation of the tangent at $(\sqrt6\cos t,\sqrt2\sin t)$:
$$\sqrt6\cos t\cdot\dfrac x6+\sqrt2\sin t\cdot\dfrac y2=1$$
So, the equation of the perpendicular through the center $(0,0)$ of the ellipse will be
$$\sqrt2\sin t\cdot\dfrac x2-\sqrt6\cos t\cdot\dfrac y6=0$$
Now solve for $x,y$ to find the coordinate of the foot.
Find $\dfrac xy,$ replace this value of $\tan t$ in
$$x=\cdots$$ which is a function of $\cos t,\sin t$