Focus of Hyperbola from Transverse and Conjugate Axis

Theorem

Let $K$ be a hyperbola whose transverse axis is $2 a$ and whose conjugate axis is $2 b$.

Let $c$ be the distance of the foci of $K$ from the center.

Then:

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

Proof

 

Let the foci of $K$ be $F_1$ and $F_2$.

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

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

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 hyperbola.

This is true for all points on $K$.

In particular, it holds true for $V_2$, for example.

Thus:

\(\ds d\)

\(=\)

\(\ds F_1 V_2 - F_2 V_2\)

\(\ds \)

\(=\)

\(\ds \paren {c + a} - \paren {c - a}\)

\(\ds \)

\(=\)

\(\ds 2 a\)

This needs considerable tedious hard slog to complete it. In particular: Some weird magic happens, and then: To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds c\)

\(=\)

\(\ds \sqrt {a^2 + b^2}\)

\(\ds \leadsto \ \ \)

\(\ds a^2 + b^2\)

\(=\)

\(\ds c^2\)

$\blacksquare$