TITLE: 20 years later need geometry IRL - Solving an ellipse to pass through 2 points. I have a line as a tangent passing through each point.
QUESTION [5 upvotes]: This is an IRL problem. I'm working in an architecture studio and trying to build an object using gdl (geometric descriptive language) code that would draw a curved ramp with different width at the top and bottom for any given angle of the ramp. In a simple case, one side of the ramp can be an arc of a circle but the other would be an ellipse. In a more complex variant, neither would be an arc of a circle. I believe this should not make a difference as it would be simply solving two equations that are independent of one another.
With given the top and bottom widths, outer radius and angle as known variables, how could I solve this?
90 degree ramp:

60 degree ramp:

REPLY [2 votes]: Place the origin where "Bottom Tangent" meets "Bottom Width" and any ellipse has equation $$\left(\frac{x-a}a\right)^2+\left(\frac y b\right)^2=1$$
You know you want this ellipse to pass through a certain point. In your second diagram it is $\Big(5000-8000\cos(60^{\circ}),8000\sin(60^{\circ})\Big)=\big(1000,4000\sqrt{3}\big)$. If you plug this in for $(x,y)$ you get a relation between $a$ and $b$.
Also taking $\frac12\frac{d}{dx}$, we get
$$\frac{x-a}{a^2}+\frac {yy'} {b^2}=0$$
Plug in the same point for $(x,y)$. Also at that location, for tangency you want $y'=\tan(30^{\circ})=\frac{1}{\sqrt{3}}$. Now you have a second relation in $a$ and $b$. You can solve the system of two equations in $a$ and $b$ and have a cartesian equation for the ellipse.

To demonstrate in the specific case of your second diagram,
$$
\left\{
\begin{aligned}
\left(\frac{1000-a}a\right)^2+\left(\frac{4000\sqrt{3}}{b}\right)^2
&=1\\
\frac{1000-a}{a^2}+\frac {4000} {b^2}
&=0
\end{aligned}\right.
$$
The second equation lets you isolate $b^2$ and sub it into the first equation to get
$$\left(\frac{1000-a}a\right)^2+\frac{12000(a-1000)}{a^2}
=1$$
$$\implies\left(1000-a\right)^2+12000(a-1000)
=a^2$$
$$\implies a=1100$$
From which:
$$\frac{1000-1100}{1100^2}+\frac {4000} {b^2}=0$$
$$\implies\frac {4000} {b^2}=\frac{100}{1100^2}$$
$$\implies b=2200\sqrt{10}$$
So with the origin where I placed it, your second diagram's ellipse is
$$\left(\frac{x-1100}{1100}\right)^2+\left(\frac{y}{2200\sqrt{10}}\right)^2=1$$
Here is that solution alongside your outer circle and $60^{\circ}$ line, plotted using GeoGebra.

If you want to generalize this beyond your second diagram, you can study you to replace the numbers ($5000$, $8000$, $60^{\circ}$, etc.) with general parameters.