Equation of Ellipse in Reduced Form/Cartesian Frame

Theorem
Let $K$ be an ellipse aligned in a cartesian coordinate plane such that:
 * the major axis of $K$ is aligned with the X-axis
 * the minor axis of $K$ is aligned with the Y-axis.

Let:
 * the major axis of $K$ have length $2a$
 * the minor axis of $K$ have length $2b$.

The equation of $K$ is:
 * $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

Proof

 * EllipseEquation.png

Let the foci of $K$ be $F_1$ and $F_2$, located at $\left({-c, 0}\right)$ and $\left({c, 0}\right)$ respectively.

Let the vertices of $K$ be $V_1$ and $V_2$, located at $\left({-a, 0}\right)$ and $\left({a, 0}\right)$ respectively.

Let the covertices of $K$ be $C_1$ and $C_2$, located at $\left({0, -b}\right)$ and $\left({0, b}\right)$ respectively.

Let $P = \left({x, y}\right)$ be an arbitrary point on the locus of $K$.

From the equidistance property of $K$ we have that:


 * $F_1 P + F_2 P = d$

where $d$ is a constant for this particular ellipse.

Applying Pythagoras' Theorem:
 * $(1): \quad \sqrt {\left({x - c}\right)^2 + y^2} + \sqrt {\left({x + c}\right)^2 + y^2} = d$

Taking the particular instance of $P$ to be $V_2 = \left({a, 0}\right)$ it follows that $d = 2 a$.

Taking the particular instance of $P$ to be $C_2 = \left({0, b)}\right)$:

Thus: