For simplicity, let's say that the ellipse is horizontal, centered at $\left(0\,0\right)$ with the following foci: E at $\left(-c\,0\right)$ and F at $\left(c\,0\right)$. Thus, the distance from each focus to the center is c.

The distance from a general point $P\left(x\,y\right)$ on the ellipse to E is given by $\mathrm{PE}equals;\sqrt{{\left(x-\left(-c\right)\right)}^{2}plus;{\left(y-0\right)}^{2}}equals;\sqrt{{\left(xplus;c\right)}^{2}plus;{y}^{2}}$.

The distance from P to F is given by $\mathrm{PF}equals;\sqrt{{\left(x-c\right)}^{2}plus;{\left(y-0\right)}^{2}}equals;\sqrt{{\left(x\mathit{}\mathit{-}\mathit{}c\right)}^{2}plus;{y}^{2}}$.

Looking at the case in which P is a vertex of the ellipse and adding up the distances from this vertex to each focus, we see that the sum of these distances is $2a$.

Thus, we know that:

$\mathrm{PE}\+\mathrm{PF}\=2a$

$\sqrt{{\left(x\+c\right)}^{2}\+{y}^{2}}\+\sqrt{{\left(x-c\right)}^{2}\+{y}^{2}}\=2a$

$\sqrt{{\left(x\+c\right)}^{2}\+{y}^{2}}equals;2a-\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

${\left(x\+c\right)}^{2}\+{y}^{2}equals;{\left(2a-\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}\right)}^{2}$

${x}^{2}\+2cx\+{c}^{2}\+{y}^{2}\=4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}\+{y}^{2}}\+{\left(x-c\right)}^{2}\+{y}^{2}$

${x}^{2}\+2cx\+{c}^{2}\+{y}^{2}equals;4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}plus;{x}^{2}-2cxplus;{c}^{2}plus;{y}^{2}$

$2cxequals;4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}-2cx$

$4cx-4{a}^{2}equals;-4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

$cx-{a}^{2}equals;-a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

${\left(cx-{a}^{2}\right)}^{2}equals;{a}^{2}\left({\left(x-c\right)}^{2}plus;{y}^{2}\right)$

${c}^{2}{x}^{2}-2{a}^{2}cx\+{a}^{4}equals;{a}^{2}{x}^{2}-2{a}^{2}cxplus;{a}^{2}{c}^{2}plus;{a}^{2}{y}^{2}$

${c}^{2}{x}^{2}\+{a}^{4}equals;{a}^{2}{x}^{2}plus;{a}^{2}{c}^{2}plus;{a}^{2}{y}^{2}$

${a}^{4}-{a}^{2}{c}^{2}equals;{a}^{2}{x}^{2}-{c}^{2}{x}^{2}plus;{a}^{2}{y}^{2}$

${a}^{2}\left({a}^{2}-{c}^{2}\right)\=\left({a}^{2}-{c}^{2}\right){x}^{2}\+{a}^{2}{y}^{2}$

Considering the case when P is the co-vertex located at $\left(0\,b\right)$, we see that ${b}^{2}plus;{c}^{2}equals;{a}^{2}$, so ${b}^{2}equals;{a}^{2}-{c}^{2}$. Substituting ${b}^{2}$, we get:

${a}^{2}{b}^{2}equals;{b}^{2}{x}^{2}plus;{a}^{2}{y}^{2}$

$\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}equals;\frac{{b}^{2}{x}^{2}plus;{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}$

$1equals;\frac{{x}^{2}}{{a}^{2}}plus;\frac{{y}^{2}}{{b}^{2}}$

This is the standard equation for an ellipse centered at $\left(0\,0\right)$ with horizontal semi-major axis a and semi-minor axis b.