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:
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
Best Answer
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 $b$,
$$(y_d-b)^2-kx_d(y_d-b)=b^2$$
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 $a$.