# Given the distance between focus points and perimeter of a ellipse, calculate the length of major and minor axis?

Focus points: F1,F2

Given: D1, distance between F1 and F2. D2, perimeter of the ellipse.

Need: D3, length of the major axis. D4, length of the minor axis.

Perfect if answered by MATLAB Code, thanks.

• Do you have any thoughts of your own at all regarding this problem? – Arthur Nov 2 '16 at 23:46

## 1 Answer

In $\mathbb R^2 ,$ with $A>B>0,$ the ellipse $\{(x,y): (x/A)^2+(y/B)^2=1\}$ has foci at $(\pm \sqrt {A^2-B^2}\;,0).$ So $D_1=2\sqrt {A^2-B^2}.$

Now the perimeter $$D_2=\int_0^{2\pi}(A^2\sin^2 t+B^2\cos^2 t)^{1/2}\;dt=\int_0^{2\pi}(B^2+(D_1/2)^2\sin^2 t)^{1/2}\;dt.$$ For a fixed value of $D_1,$ the integral is a strictly increasing function of $|B|,$ so there is at most one value $B>0$ for which the integral equals $D_2.$ If we know $B$ we know $A,$ because $A=\sqrt {D_1^2+B^2}\;.$ The semi-major and semi-minor axes' lengths are $2A$ and $2B.$

But the integral cannot be simplified.