= Solution
Substituting $\omega=U\omega'$ and using the diagonalization gives
$$
\boxed{T'(\omega_x',\omega_y')=\frac{I_0}{2}\omega_x'^2+\frac{3I_0}{2}\omega_y'^2.}
$$
For $I_0>0$ and energy level one in the chosen units, its <ellipse> equation is
$$
\frac{\omega_x'^2}{2/I_0}+\frac{\omega_y'^2}{2/(3I_0)}=1.
$$
The major semiaxis has length $a=\sqrt{2/I_0}$ along $u_1=(1,1)/\sqrt2$; the minor semiaxis has length $b=\sqrt{2/(3I_0)}$ along $u_2=(-1,1)/\sqrt2$. The full axis lengths are $2a$ and $2b$. The original-coordinate curve is the same ellipse tilted by $45^\circ$, elongated along $\omega_y=\omega_x$; an orthogonal coordinate change preserves lengths and does not change the ellipse itself.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017-rotational-energy-ellipse.png]
{title=Principal axes of the rotational-energy ellipse}
{description=Original plot at I0 equal to one and energy equal to one, showing the longer axis along the normalized (1,1) eigenvector and the shorter perpendicular axis.}
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