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.... 

 A: 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.
A: Partial answer, based on the fact that the image of any ellipse under affine transformation is another ellipse.
Let the equation of the ellipse be $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ or $$b^2 x^2 + a^2 y^2 = 1$$ and let the desired semi-diameter intersect the ellipse at $(h, k)$ where $h^2 + k^2 = r^2$.
I claim that there is an affine transformation consisting solely of a scaling along the semi-diameter that can transform this ellipse into a circle. Specifically, write $T = PAP^{-1}$, where $$P = \begin{bmatrix}h & -k \\ k & h \end{bmatrix}$$ and $$P^{-1} = \dfrac{1}{r^2} \begin{bmatrix} h & k \\ -k & h \end{bmatrix}$$ transform coordinates between the standard basis and the basis vectors $(h, k)$ and $(k, -h)$, and $$A = \begin{bmatrix} \lambda & 0 \\ 0 & 1 \end{bmatrix}.$$
As $\det T = \lambda$ and $T$ preserves distances from the axis of rotation, the rotational inertia of the resulting shape is $\lambda$ times that of the original shape. The matrix expression for $T$ is $$T = \frac{1}{r^2} \begin{bmatrix} \lambda h^2 + k^2 & hk - \lambda h k \\ hk - \lambda hk & h^2 + \lambda k^2 \end{bmatrix}$$ and the squared radius $R(x, y) := ||T(x, y)||^2$ of the image of some point $(x, y)$ is thus (omitting a bunch of tedious steps in the algebra) \begin{align*}
R &= \frac{1}{r^4} \left[  (\lambda h^2 + k^2) x + (hk - \lambda hk) y \right]^2 + \frac{1}{r^4} \left[(hk - \lambda hk) x + (hk - \lambda hk) y \right]^2 \\
&= \frac{1}{r^2} \left( \lambda^2 (hx - ky)^2 + (hy + kx))^2 \right) 
\end{align*}
and finding the value of $\lambda$ for which this expression is constant (and the radius of the corresponding circle) is just a straightforward tedious exercise in algebra or differential calculus from which the final answer would follow readily.
