Let's say I have a large ellipse on a flat ground as shown in Figure 1 and 2 below. Angles DBC and ABD are right-angles and the longest radius with respect to the ground, BD is perpendicular to line ABC and the flat ground. The ellipse is in a stationery state in Figure 1.
Let's imagine I tilt the ellipse at point C as shown in Figure 2 until point C is 0.3 meters above the flat ground (or, CE is 0.3 meters).
How do I find the length of AF? Note that AF and CE are perpendicular to the flat ground.
Let us rotate the second figure so that $AC$ is horizontal. Now $EF$ is a tangent line to the lower half of the ellipse.
At this point the equation of the ellipse is:
$$\left(\dfrac{x}{30}\right)^2 + \left(\dfrac{y}{1.5}\right)^2 = 1 $$
And the function for the lower half of it is:
$$y = -1.5\sqrt{1 - \left(\dfrac{x}{30}\right)^2} $$
Recall that the equation of the tangent line to a function $f(x)$ through the point $(x_0,f(x_0))$ is
$$ y - f(x_0) = f'(x_0)(x - x_0)$$
With our curve, the equation of a tangent to the lower half of the ellipse is:
$$y + 1.5\sqrt{1 - \left(\dfrac{x_0}{30}\right)^2} = \dfrac{x_0}{20\sqrt{30^2 - x_0^2}}(x - x_0) $$
or, equivalently (after some algebraic manipulation)
$$y = \left(\dfrac{x_0}{20\sqrt{30^2 - x_0^2}}\right)x - \left(\dfrac{60^2}{20\sqrt{30^2 - x_0^2}}\right)$$
Now we have to find $x_0$ so that this line is $EF$, i.e. the distance between the line and $C$ is $0.3$.
The distance between a line $y = ax + b$ and a point $(X,Y)$ is
$$\dfrac{|aX-Y + b|}{\sqrt{a^2 +1}} $$
If our axis are the same as the ellipse axis, then our point $C$ has coordinates $(30,0)$ so our distance is
$$\dfrac{\left|\dfrac{30x_0 - 30^2}{20\sqrt{30^2 - x_0^2}}\right|}{\dfrac{\sqrt{600^2 - (20^2 - 1)x_0^2}}{20\sqrt{30^2 - x_0^2}}} = \dfrac{900 - 30x_0}{\sqrt{360000 - 399x_0^2}}$$
We have to find $x_0$ so that this number is $0.3$:
$$0.3 = \dfrac{900 - 30x_0}{\sqrt{360000 - 399x_0^2}} \implies \sqrt{360000 - 399x_0^2} = 3000 - 100x_0 \implies $$ $$360000 - 399x_0^2 = 9000000 - 600000x_0 + 10000x_0^2 \implies $$ $$10399x_0^2 - 600000 x_0 + 8640000 = 0$$
The solutions are
$$x_0 = \dfrac{300000 \pm 1200\sqrt{106}}{10399}$$
Since $x_0 \in [-30,30]$ we only consider the solution with the "minus". This gives us the equation for $EF$.
Now we have to find the tangent line to the ellipse parallel to $CE$. Out of the two that exist, we will select the one closest to $A$. Since that line is perpendicular to $EF$, then the slope of said line is
$$-\dfrac{20\sqrt{30^2 - x_0^2}}{x_0} = -\sqrt{\dfrac{2503+250\sqrt{106}}{72}}$$
So for some $x_1$ we have that
$$\dfrac{x_1}{20\sqrt{30^2 - x_1^2}} = -\sqrt{\dfrac{2503+250\sqrt{106}}{72}}$$
therefore
$$\dfrac{x_1^2}{12000 -400 x_1^2} = \dfrac{2503+250\sqrt{106}}{72} \implies $$ $$ (1001272 + 100000\sqrt{106})x_1^2 = 12000(2503 + 250\sqrt{106})$$
Solving for $x_1$ we get
$$x_1 = -\sqrt{\dfrac{2996170500 - 375000\sqrt{106}}{99747191}}$$
Plugging this into our ellipse equation we get $y_1$:
$$y_1 = -\sqrt{\dfrac{867763014 + 3750\sqrt{106}}{398988764}}$$
Finally, plugging this points into our point-line distance equation we get that the distance needed is:
$$\dfrac{|ax_1 - y_1 + b|}{\sqrt{a^2 + 1}}$$
Where $$a = \sqrt{\dfrac{72}{2503+250\sqrt{106}}} \qquad b = \dfrac{3600a}{x_0}$$
Numerically, the solution is approximately $13.2781237$ meters.