I have two ellipses which are both not rotated, but have different sizes. I want to know if the smaller ellipse 1 is fully inside the larger ellipse 0. In my special case both ellipses have the same axis ratio, so $a_0/b_0=a_1/b_1$, but I doubt this makes a difference for the calculation.
I have started to work on a solution, but I am not so sure if this is the right approach and if it will ultimately lead to success. Here is what I have come up with so far:
The radius of ellipse 0 can be calculated for any point on the ellipse as
$$ r_0=\sqrt{(x-x_0)^2+(y-y_0)^2} $$ The general equation for an ellipse gives a relation between $x$ and $y$ for points on the ellipse.
$$ \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1 $$
$$ (y-y_0)^2=b*\sqrt{1-\frac{(x-x_0)^2}{a^2}} $$
This can be used to calculate radius of both ellipses depending on the x coordinate.
$$ r_0 = \sqrt{(x-x_0) + b_0^2(1-\frac{(x-x_0)^2}{a_0^2})} $$
$$ r_1 = \sqrt{(x-x_1) + b_1^2(1-\frac{(x-x_1)^2}{a_1^2})} $$
To calculate the distance $s$ between both ellipses at a certain $x$ coordinate, we simply have to add the distance $d_{01}$ between the center points to radius 1 and subtract this from radius 0.
$$ s=r0 - (d_{01} + r1) $$ where $$ d_{01} = \sqrt{(x_1-x_0)^2 + (y1-x_0)^2} $$
Now I have an equation where $s$ only depends on $x$. My idea is to get the minimum of $s$ through the first derivative. If the minimum is non-negative, ellipse 1 is inside or on ellipse 0. If it is negative, part of ellipse 1 is outside ellipse 0.
Here are my questions:
- Am I on the right way so far?
- Is calculating the minimum for s the right approach?
- Do I have a chance to actually calculate the minimum of $s$?
Note: Ther is a similar question here, but I think my case is much simpler because the ellipses are not rotated similar.

Because they have the same axis ratio and are orientated the same, the easy solution is to multiply all y-coordinates by $\frac{a}{b}$. Then you have two circles.
All points on the large circle are within $a_0$ of its centre whereas all points on the small circle are within $a_l+\sqrt{(x_l-x_0)^2+\frac{a^2}{b^2}(y_l-y_0)^2}$.