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nykkyo · Jul 9, 2011, 02:19 AM

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

jcaron2 · Jul 9, 2011, 08:51 PM

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.

nykkyo · Jul 9, 2011, 09:59 PM

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