The locus of the foot of the perpendicular from a focus to a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is the circle $x^2 + y^2 = a^2$
I am able to solve it using tangent as $y=mx±\sqrt{a^2m^2+b^2}$ but my student wants me to solve it using $x=a\cos \theta$ and $y=b\sin \theta$. I am not able to solve it.
Using this, the equation at $P(a\cos t,b\sin t)$
is $$xb\cos t+y a\sin t-ab=0$$
Method$\#1:$
Now if $R(h,k)$ is the foot of the perpendicular from focus $A(ae,0)$
$$hb\cos t+k a\sin t-ab=0\ \ \ \ (1)$$
and as $PR$ will be perpendicular to the tangent, form the second equation $$\dfrac{0-k}{ae-h}\cdot-\dfrac{b\cos t}{a\sin t}=-1\ \ \ \ (2)$$
Solve for $h,k$ and eliminate $t$
Method$\#2:$
As the gradient of the tangent is $$-\dfrac{b\cos t}{a\sin t},$$ the equation of the perpendicular from focus $A(ae,0)$ will be
$$\dfrac{a\sin t}{b\cos t}=\dfrac{y-0}{x-ae}$$
Find the intersection of the perpendicular with the tangent and eliminate $t$