Height of a rotated ellipse If I have an ellipse, it is easy to find its height, twice the length of the major axis. But if the ellipse is rotated a certain number of degrees, how do you find the vertical height from top to bottom?
 A: Hint:
An ellipse of center in the origin and the axis rotated by an angle $\theta$ has equation:
$$
\frac{(x\cos \theta+y\sin \theta)^2}{a^2}+\frac{(y\cos \theta-x\sin \theta)^2}{b^2}=1
$$
that can be write as:
$$
Ax^2+Bxy+Cy^2=1
$$
with $B^2-4AC<0$.
From this find:
$$
y=\dfrac{-Bx\pm\sqrt{B^2x^2-4C(Ax^2-1)}}{2C}
$$
and you have two equation of two semi-ellipses. Now find the maximum and minimum of these functions and the difference of the ordinates of these points is the searched height.
A: The equation of a rotated ellipse is:
$$
\frac{(x\cos \theta+y\sin \theta)^2}{a^2}+\frac{(-x\sin \theta+y\cos \theta)^2}{b^2}=1
$$
Expand the equation:
$$
\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)x^2+2\cos\theta\sin\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)xy+\left(\frac{\sin^2\theta}{a^2}+\frac{\cos^2\theta}{b^2}\right)y^2=1
$$
This can be written as $Ax^2+Bxy+Cy^2=1$. Solve for x:
$$
x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}
$$
The maximum and minimum values of $y$ are where there is only one value of $x$, i.e. where $B^2-4AC=0$, so we solve that equation:
$$
\left(2\cos\theta\sin\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)y-1\right)^2-4\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)\left(\frac{\sin^2\theta}{a^2}+\frac{\cos^2\theta}{b^2}\right)y^2=0
$$
$$
a^2+b^2-2y^2-(a^2-b^2)\cos2\theta=0
$$
$$
y=\pm\sqrt{\frac{1}{2}\left(a^2+b^2-(a^2-b^2)\cos2\theta\right)}
$$
Finally, the height of the ellipse is twice this distance, $2y$:
$$
\sqrt{2\left(a^2+b^2-(a^2-b^2)\cos2\theta\right)}
$$
A: Hint: The height is expressed by y. Extract y in terms of x, using the equation of the rotated ellipse, and then remember what you were taught in school about finding the maximum and minimum of a function...
