How can you prove that the sum of radii from the foci to the ekkipse is constant?

By choosing the correct coordinate system, any ellipse can be written in canonical form:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

Written in this form, the foci of the ellipse are at the points

(\pm \sqrt{a^2-b^2},0)

Now we need to find the distance from a point on the ellipse to each focus. First, let's rewrite the equation in the form of

y = \pm b\sqrt{1-\frac{x^2}{a^2}}

Therefore a point on the ellipse can be written as

p = (x,\pm b\sqrt{1-\frac{x^2}{a^2}})

Now we need to find the distance between the foci and p.

d_{1,2} = \sqrt{(x\mp\sqrt{a^2-b^2})^2+b^2(1-\frac{x^2}{a^2})}

Expanding then simplifying, we get

d_{1,2} = \sqrt{x^2 \mp 2x\sqrt{a^2-b^2}+a^2-b^2+b^2-b^2(\frac{x^2}{a^2})}

d_{1,2} = \frac 1a\sqrt{(a^2-b^2)x^2 \mp 2a^2x\sqrt{a^2-b^2}+a^4}

d_{1,2} = \frac 1a\sqrt{(a^2 \mp x\sqrt{a^2-b^2})^2}

d_{1,2} = a \mp \frac xa\sqrt{a^2-b^2}

d_1 + d_2= a-\frac xa\sqrt{a^2-b^2}+a+\frac xa\sqrt{a^2-b^2}

d_1 + d_2= 2a

Note that the sum of the distances is constant, independent of x.

Thanks. I also found the result on wikiredi, using polar coordinates.