I am trying to prove that for the ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ the sum of the distances from the gernal point to the foci ( $(ae,0)$ and $(-ae,0)$ were $e$ is the eccentricity) is always $2a$, I have tired using Pythagorus and the parmetric equations of the ellipse but this just got some horible algebra that seemed impossible to solve. Can you please give me some hints of how to tackle it, thanks?


1 Answer 1


The sum of distances you want to estimate is given by

$$ d = \sqrt{(x_p + a e)^{2} + y_p^{2}} + \sqrt{(x_p - a e)^{2} + y_p^{2}} $$

where $(x_p,y_p)$ is a given point on the ellipse. Since $(x_p,y_p)$ is on the ellipse, you want to replace $y_p^{2}$ in $d$, by a function of $x_p$ in order to have $d$ expressed only as a function of $x_p$. The condition $(x_p / a )^{2} + (y_p / a)^{2} = 1 $ allow us to write

$$ y_p^{2} = (1 - e^{2}) \, a^{2} - (1 - e^{2}) \, x_{p}^{2}$$

Injecting this in the expression of $d$, we obtain after simplification (you might want to check it yourself !)

$$ \begin{aligned} d \; & = \sqrt{(e x_{p} - a)^{2}} + \sqrt{(e x_{p} + a)^{2}} \\ & = \left| e x_{p} - a \right| + \left| e x_{p} + a \right| \end{aligned}$$

The final step of the calculation is to remember that we have $|x_{p}| \leq a $ and $e \leq 1$, so that we may rewrite the absolute values as

$$\begin{aligned} d \; & = (a - e x_{p}) + (a + e x_{p}) \\ & = 2 a \end{aligned}$$


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