I've been looking at this for hours now, trying to solve it using the ellipse equation and looking at other questions here, unfortunately with no success.

Here is what I am trying to achieve:

enter image description here

I have:

  • big axis of ellipse is parallel to x axis
  • ellipse is tangent to x axis in origin (= point A)
  • the position of point D
  • the slope of the tangent to the ellipse in D

I am looking for:

  • length of major axis of the ellipse (EF)
  • length of minor axis of the ellipse (AG)

How would you approach this? I feel like there should be only one solution to this problem and that it is well defined, but I could not solve it...

Thanks a lot for your help


Let $D(x_d,y_d)$ and $k$ the slope of the tangent. The ellipse has the form,

$$\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$$

Its derivative is $y'=-\frac{b^2x}{a^2(y-b)}$ and matches $k$ at D,

$$k = -\frac{b^2x_d}{a^2(y_d-b)} \tag{1}$$

The point $(x_d,y_d)$ satisfies,

$$\frac{x_d^2}{a^2} + \frac{(y_d-b)^2}{b^2} = 1\tag{2}$$

Combine (1) an (2) to get the following equation for $b$,


which yields the unique solution for the minor axis $b$,

$$b= \frac{y_d(y_d-kx_d)}{2y_d-kx_d}$$

Then, plug the solution $b$ into (1) to get the major axis $a$.

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  • $\begingroup$ It might be worth noting that besides the obvious condition $2d_y-kx_d=0$, there’s also no such ellipse when $b\le0$. $\endgroup$ – amd Oct 9 '19 at 20:13
  • $\begingroup$ Thanks, I was missing multiple steps indeed! $\endgroup$ – zigoingoin Oct 10 '19 at 18:50

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