# Equation of Ellipse in Reduced Form/Cartesian Frame/Proof 1

## Theorem

Let $K$ be an ellipse aligned in a cartesian plane in reduced form.

Let:

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

The equation of $K$ is:

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

## Proof By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.

Let the vertices of $K$ be $V_1$ and $V_2$.

By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.

Let the covertices of $K$ be $C_1$ and $C_2$.

By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.

Let $P = \tuple {x, y}$ 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.

$d = 2 a$

Also, from Focus of Ellipse from Major and Minor Axis:

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

Then:

 $\ds \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}$ $=$ $\ds d = 2 a$ Pythagoras's Theorem $\ds \leadsto \ \$ $\ds \sqrt {\paren {x + c}^2 + y^2}$ $=$ $\ds 2 a - \sqrt {\paren {x - c}^2 + y^2}$ $\ds \leadsto \ \$ $\ds \paren {x + c}^2 + y^2$ $=$ $\ds \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2$ squaring both sides $\ds \leadsto \ \$ $\ds x^2 + 2 c x + c^2 + y^2$ $=$ $\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2$ expanding $\ds \leadsto \ \$ $\ds x^2 + 2 c x + c^2 + y^2$ $=$ $\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2$ further expanding $\ds \leadsto \ \$ $\ds a^2 - c x$ $=$ $\ds a \sqrt {\paren {x - c}^2 + y^2}$ gathering terms and simplifying $\ds \leadsto \ \$ $\ds \paren {a^2 - c x}^2$ $=$ $\ds a^2 \paren {\paren {x - c}^2 + y^2}^2$ squaring both sides $\ds \leadsto \ \$ $\ds c^2 x^2 - 2 c x a^2 + a^4$ $=$ $\ds a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2$ expanding $\ds \leadsto \ \$ $\ds c^2 x^2 + a^4$ $=$ $\ds a^2 x^2 + a^2 c^2 + a^2 y^2$ simplifying $\ds \leadsto \ \$ $\ds a^4 - a^2 c^2$ $=$ $\ds a^2 x^2 - c^2 x^2 + a^2 y^2$ gathering terms $\ds \leadsto \ \$ $\ds a^2 \paren {a^2 - c^2}$ $=$ $\ds \paren {a^2 - c^2} x^2 + a^2 y^2$ simplifying $\ds \leadsto \ \$ $\ds a^2 b^2$ $=$ $\ds b^2 x^2 + a^2 y^2$ substituting $a^2 - c^2 = b^2$ from $(2)$ $\ds \leadsto \ \$ $\ds 1$ $=$ $\ds \frac {x^2} {a^2} + \frac {y^2} {b^2}$ dividing by $a^2 b^2$

$\blacksquare$