Equation of Hyperbola in Reduced Form/Cartesian Frame/Proof 1

Proof

 * HyperbolaReducedForm.png

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.

From Equidistance of Hyperbola equals Transverse Axis:
 * $d = 2 a$

Also, from Focus of Hyperbola from Transverse and Conjugate Axis:
 * $c^2 a^2 = b^2$

, let us choose a point $P$ such that $F_1 P > F_2 P$.

Then: