# Equation of Hyperbola in Reduced Form/Cartesian Frame

## Theorem

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

Let:

the transverse axis of $K$ have length $2 a$
the conjugate 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:

$\size {F_1 P - F_2 P} = d$

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

$d = 2 a$
$c^2 a^2 = b^2$

Without loss of generality, let us choose a point $P$ such that $F_1 P > F_2 P$.

Then:

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

$\blacksquare$