Finding Moment of Inertia of ellipse. Show that the moment if inertia of an elliptic area of mass 
M
and semi-axis 
a
and 
b
about a semi-diameter 
of length 
r
is $$\frac{Ma^2b^2}{4r^2}$$.
My attempt.


*

*I know that MI about ox is ${Mb^2 \over 4}$.

*MI about oy axis is ${Ma^2 \over 4}$.
How to proceed further.... 


Let $\theta$ be the angle between the axis you wish to compute MI and the $x$-axis.
For any point $(x,y)$ in the plane, its distance to the axis equals to $|x\sin\theta - y\cos\theta|$. The desired MI is given by following integral
$${\rm MI}_\theta \stackrel{def}{=} \rho\int_{\Omega} (x\sin\theta - y\cos\theta)^2 dx dy$$
where $\rho$ is mass density and $\Omega$ is the region for the ellipse.
Expand the integrand and notice under transform $y \mapsto -y$, $\Omega$ remains invariant while the cross term proportional to $xy$ pickup a minus sign. The cross term will not contribute anything to the integral. This leads to
$$MI_\theta = \left(\rho \int_{\Omega} x^2 dxdy\right) \sin^2\theta + \left(\rho\int_{\Omega} y^2 dx dy\right) \cos^2\theta$$
The two coefficients inside the parentheses are nothing but the MI with respect to $y$ and $x$-axes. This means
$$MI_{\theta} = \frac{M}{4}((a\sin\theta)^2 + (b\cos\theta)^2)
= \frac{Ma^2b^2}{4}\left(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\right)
$$
Since the axis has semi-diameter $r$, when you start at origin and move forward for a distance $r$, you will hit the circumference of ellipse. This implies
$$\frac{(r\cos\theta)^2}{a^2} + \frac{(r\sin\theta)^2}{b^2} = 1 \quad\implies\quad
\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} = \frac{1}{r^2}$$
Substitute this back into above expression of ${\rm MI}_\theta$ and you are done.
