How high is the highest point ?

The simple way to go about solving this problem is using the fact that an ellipse with semi-axes $a , b$ that is tangent to both the $x$ and $y$ axes, has its center $C=(C_x, C_y)$ satisfying

$C_x^2 + C_y^2 = a^2 + b^2$

Now, from symmetry, we know that $C_x = \dfrac{30}{2} = 15$ , hence $C_y = \sqrt{ 100 + 400 - 225 } = \sqrt{ 275 }$ . And using the symmetry argument again, it follows that the highest point $y$ -coordinate is $H = 2 C_y = 2 \sqrt{275} = 33.166$

Below is a formula for a generalized 2D ellipse. The ellipse center is at $(x_0,y_0)$ . Its semi-major and semi-minor axis lengths are $a$ and $b$ , and the major axis is oriented at an angle $\theta$ with respect to the positive $x$ axis.

$A x^2 + B x y + C y^2 + D x + E y + F = 0 \\ A = a^2 \, \sin^2 \theta + b^2 \, \cos^2 \theta \\ B = 2(b^2 - a^2) \, \sin \theta \, \cos \theta \\ C = a^2 \, \cos^2 \theta + b^2 \, \sin^2 \theta \\ D = -2 A x_0 - B y_0 \\ E = -B x_0 - 2 C y_0 \\ F = A x_0^2 + B x_0 y_0 + C y_0^2 - a^2 b^2$

When $x = 0$ , and assuming that there is only one corresponding $y$ value:

$C y^2 + E y + F = 0 \\ \implies E^2 - 4 C F = 0$

When $y = 0$ , and assuming that there is only one corresponding $x$ value:

$A x^2 + D x + F = 0 \\ \implies D^2 - 4 A F = 0$

When $x = x' = 30$ , and assuming that there is only one corresponding $y$ value:

$A x'^2 + B x' y + C y^2 + D x' + E y + F = 0 \\ \implies (B x' + E)^2 - 4 C (A x'^2 + D x' + F) = 0$

After numerical solution, the values which satisfy these equations are:

$x_0 \approx 15.000 \\ y_0 \approx 16.583 \\ \theta \approx 49.797^\circ$

The highest point occurs at $y \approx 33.166$ .