Showing that the distance from any point on an ellipse to the foci points is constant Let $a$ be the largest x-value of an ellipse centered at the origin. Similarly, let $b$ be the largest y-value of the ellipse. Assume that $a>b$. We know that the foci points will be at $c_1=(-\sqrt {a^2-b^2},0)$ and $c_2=(+\sqrt {a^2-b^2},0)$ respectively. Then select some point on the ellipse $p_1$ with x-value $x_1$ between $-a$ and $a$. If we only consider the positive y-values of the ellipse, then we know from the equation of an ellipse that the y-value of $p_1$, will be $y_1=\sqrt{b^2(1-\frac{x_1^2}{a^2}})$ or better yet $y_1=\frac{b\sqrt{a^2-x_1^2}}{a}$. So for a generic x-value, the point on the ellipse will be $p_1=\left(x_1, \frac{b\sqrt{a^2-x_1^2}}{a}\right)$.
I'd like to show that the sum distance of $p_1$ to $c_1$ and $p_1$ to $c_2$ will be a constant that does not depend on the value of $x_1$. To do so, I planned to use Euclid's Metric, and have cancellation of the $x_1$ terms algebraically. 
In other words, I would like to show a formula in terms of only $a$ and $b$ for the distance of a point on the ellipse to each foci. I am looking for how to manipulate Euclid's metric to show this cancellation.
For thoroughness, I'll give the original equation for the generic distance from $p_1$ to $c_1$: $\sqrt{(x_1-\sqrt{a^2-b^2})^2+\left(\frac{b\sqrt{a^2-x_1^2}}{a}\right)^2}$. 

Every point of elipse $(bx)^2 +(ay)^2 =(ab)^2 $ can be represent as $P=(a\cos t , b\sin t)$ hence $$c_1 P +c_2 P =\sqrt{(\sqrt{a^2 -b^2} +a\cos t )^2 + b^2\sin^2 t } +\sqrt{(\sqrt{a^2 -b^2} -a\cos t )^2 + b^2\sin^2 t }=\sqrt{(a+\sqrt{a^2 -b^2}\cos t)^2 } +\sqrt{(a-\sqrt{a^2 -b^2}\cos t)^2 } =2a$$
