Relationship between two rotated ellipses centered at the origin Let there be two ellipses centered at the origin, and rotated angles $\alpha$ and $\alpha'$ from the x-coordinate axis. $a, b$ are the semiaxes of the first, $a',b'$ the semiaxes of the second, and the ratio $ab/a'b'$ is called $f$.
The equations for both ellipses can be written:
$$
\frac{(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac{(y\cos\alpha-x\sin\alpha)^2}{b^2}=1
$$
and
$$
\frac{(x\cos\alpha+fy\sin\alpha)^2}{a^2}+\frac{(fy\cos\alpha-x\sin\alpha)^2}{b^2}=1
$$
Where the equation for the second ellipse is written for convenience as a function of the parameters $a,b, \alpha$.
Now in the text I am following I find that the semiaxes of the second ellipse, $a', b'$, can be obtained from the second equation above, using the relation:
$$
\frac{1}{a'^2}+\frac{1}{b'^2}=\frac{\cos^2\alpha+f^2\sin^2\alpha}{a^2}+\frac{\sin^2\alpha+f^2\cos^2\alpha}{b^2}
$$
However I don't have any clue were this last relation comes from. Any help is welcome!
 A: The equation of an ellipse in standard position with semi-axis lengths $a'$ and $b'$ can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\frac1{a'^2}&0\\0&\frac1{b'^2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1.\tag1$$ The trace of the central matrix—the sum of its main diagonal elements—in this equation is equal to $\frac1{a'^2}+\frac1{b'^2}$. Rotating this ellipse amounts to a similarity transformation of the coefficient matrix, which leaves its trace unchanged. Expanding the equation of the rotated ellipse given in the question and collecting terms gives $$\left({\cos^2\alpha\over a^2} + {\sin^2\alpha\over b^2}\right)x^2+2f\cos\alpha\sin\alpha\left(\frac1{a^2} - \frac1{b^2}\right)xy + \left({f^2\sin^2\alpha\over a^2} + {f^2\cos^2\alpha\over b^2} \right)y^2=1.\tag2$$ In matrix form, the diagonal elements are the coefficients of $x^2$ and $y^2$, so take these two coefficients from equation (2) and rearrange their sum to get $$\frac1{a'^2}+\frac1{b'^2}={\cos^2\alpha+f^2\sin^2\alpha\over a^2}+{\sin^2\alpha+f^2\cos^2\alpha\over b^2}$$ as required.
