# Focus of Hyperbola from Transverse and Conjugate Axis

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## 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. |

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

\(\ds \leadsto \ \ \) | \(\ds a^2 + b^2\) | \(=\) | \(\ds c^2\) |

$\blacksquare$