I have an ellipse. I know that $r_1 = 28$, $r_2 = 20$.
I'm trying to find the width and height of the ellipse. I know how to find the width of the ellipse with $r_1+r_2=2a$, but I can't figure out how to find the height of the ellipse.
How do I do this?
Assume without loss of generality that $r_1 \ge r_2 > 0$, and let the equation of the ellipse in standard form be $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a \ge b > 0.$$ Then the distance from the center to a focus is given by $c$ where $c^2 = a^2 - b^2$. By the triangle inequality, we must have $r_2 + 2c \ge r_1$, which in turn implies $c \ge \frac{r_1 - r_2}{2}$. Consequently, $$b = \sqrt{a^2 - c^2} \le \sqrt{\left(\frac{r_1 + r_2}{2}\right)^2 - \left(\frac{r_1 - r_2}{2}\right)^2} = \sqrt{r_1 r_2}.$$ As such, for a general choice of $r_1, r_2$, there is an infinite family of ellipses satisfying the criteria. In particular, for a chosen $b$ in the interval $(0, \sqrt{r_1, r_2})$, the points $$(x,y) = \left(\pm \frac{r_1^2 - r_2^2}{4c}, \pm \frac{\sqrt{(2c-r_1-r_2)(2c-r_1+r_2)(2c+r_1-r_2)(2c+r_1+r_2)}}{4c} \right)$$ where again $c^2 = a^2 - b^2$, will satisfy the required distance criteria.
In your case, $r_1 = 28$, $r_2 = 20$ admits, for instance, $$\{(a,b,c), (x,y)\} = \{(24,4\sqrt{35},4), (24,0)\}$$ which is easily verified, but $$\{(a,b,c), (x,y)\} = \{(24,8,16\sqrt{2}), (3\sqrt{2}, \sqrt{62})\},$$ also works. And here is an animation showing the family of ellipses and the point on the ellipse that satisfies the given criteria:
You can see that what is really happening here is that the locus of the intersection points of two circles with radii $r_1$ and $r_2$, as their centers move along the line passing through both centers, will necessarily be a point on the ellipse, so it is easy to see that the ellipse is not uniquely determined if only these distances are given, since the circles are free to move.