Equations and Graphs of Ellipses

This ellipse has its center at the origin, foci F(c, 0) and  F′(–c, 0), and vertices V(a, 0) and V′ (–a, 0). The distance from V to F is a  −c and the distance from V to F′ is a + c. The sum of these distances is 2a. Since V is on the ellipse, this sum is the constant referred to in the definition of an ellipse.

Thus, for any point P(x, y) on the ellipse, $d\left(P,F\right)+d\left(P,F'\right)=2a$.

By the distance formula, $d\left(F,P\right)=\sqrt[]{\left(x-c\right)^2+y^2}$, and $d\left(P,F'\right)=\sqrt[]{\left[x-\left(-c\right)\right]^2+y^2}$.

Thus we have: $d\left(F,P\right)+d\left(P,F'\right)=\sqrt[]{\left(x-c\right)^2+y^2}+\sqrt[]{\left(x+c\right)^2+y^2}=2a$

Isolating $\sqrt[]{\left(x-c\right)^2+y^2}$ gives:

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

Now carefully squaring both sides we get:

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

Next, square both $x-c$ and $x+c$.

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

Isolate the square root term, $4a\sqrt[]{\left(x+c\right)^2+y^2}$.

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

Now divide both sides by 4.

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

Square each side.

$a^2\left(x^2+2cx+c^2+y^2\right)=a^4+2ca^2x+c^2x^2$

Simplify the left side by using the distributive property.

$a^2x^2+2ca^2x+a^2c^2+a^2y^2=a^4+2ca^2x+c^2x^2$

Subtract $2ca^2x$, rearrange terms, then factor.

$a^2x^2+a^2c^2+a^2y^2=a^4+c^2x^2$

$a^2x^2+c^2x^2+a^2y^2=a^4+a^2c^2$

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

Divide by $a^2\left(a^2-c^2\right)$.

$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$

Returning to this figure, since B(0, b) is on the ellipse, we have d(B, F) + d(B, F') = 2$a$.

$\sqrt[]{\left(-c\right)^2+b^2}+\sqrt[]{c^2+b^2}=2a$

$2\sqrt[]{c^2+b^2}=2a$

$\sqrt[]{c^2+b^2}=a$

$c^2+b^2=a^2$

$b^2=a^2-c^2$

Replacing $a^2-c^2\:$with $b^2$ in the equation $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$ gives the standard form of the equation of an ellipse centered at the origin with foci on the x-axis.

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

If the vertices and foci were on the y-axis, an almost identical derivation could be used to get the following standard form.

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