Fancy Rectangle

We are required to find the locus of the point $(h,k)$ from which two mutually perpendicular lines can be drawn that are normal to the ellipse.

Normal to the ellipse at the point $(a\cos\theta,b\sin\theta)$ is given by $ax\sec\theta -by\csc\theta =a^2-b^2$ and the slope of this normal is given by $m=\dfrac{a}{b}\tan\theta$

Putting $x=h, y=k$ and eliminating $\theta$ , we get a quartic equation in $m$

$b^2h^2m^4-2b^2hkm^3+(a^2h^2+b^2k^2-a^4e^4)m^2-2a^2hkm+a^2k^2=0$

This tells us that at most four normals can be drawn from the point $(h,k)$ to the ellipse whose slopes are the solutions of the above equation.

Let $m_1,m_2,m_3,m_4$ be the four solutions, and for our locus, let $m_1$ and $m_2$ be the slopes of the mutually perpendicular normals.

From theory of equations we get:

$m_1+m_2+m_3+m_4=\dfrac{2k}{h}$

$m_1 m_2+m_1 m_3+m_1 m_4+m_2 m_3+m_2 m_4+m_3 m_4=\dfrac{a^2h^2+b^2k^2-a^4e^4}{b^2h^2}$

$m_1 m_2 m_3+m_1 m_2 m_4+m_1 m_3 m_4+m_2 m_3 m_4=\dfrac{2a^2k}{b^2h}$

$m_1 m_2 m_3 m_4=\dfrac{a^2k^2}{b^2h^2}$

and $m_1 m_2=-1$

Eliminating $m_1,m_2,m_3,m_4$ from the above five equations (a very tedious yet fun exercise) , we get

$(a^2+b^2)(h^2+k^2)(a^2k^2+b^2h^2)^2=(a^2-b^2)^2(a^2k^2-b^2h^2)^2$

This is the required locus. Converting this to polar form, $h=r\cos\theta$ , $k=r\sin\theta$ , we get

$r=\dfrac{a^2-b^2}{\sqrt{a^2+b^2}}\left(\dfrac{a^2\sin^2\theta-b^2\cos^2\theta}{a^2\sin^2\theta+b^2\cos^2\theta}\right)$

Now, the required area is $\displaystyle \frac{1}{2}\int_0^{2\pi}r^2\,d\theta$

Which turns out to be $\boxed{(a-b)^2\pi}$