Equidistance of Hyperbola equals Transverse Axis
Jump to navigation
Jump to search
Theorem
Let $K$ be an hyperbola whose foci are $F_1$ and $F_2$.
Let $P$ be an arbitrary point on $K$.
Let $d$ be the constant distance such that:
- $\left\lvert{d_1 - d_2}\right\rvert = d$
where:
- $d_1 = P F_1$
- $d_2 = P F_2$
Then $d$ is equal to the transverse axis of $K$.
Proof
By the equidistance property of $K$:
- $\left\lvert{d_1 - d_2}\right\rvert = d$
applies to all points $P$ on $K$.
Thus it also applies to the two vertices $V_1$ and $V_2$.
Observing the signs of $\left\lvert{d_1 - d_2}\right\rvert$ as appropriate:
- $V_1 F_2 - V_1 F_1 = d$
- $V_2 F_1 - V_2 F_2 = d$
Adding:
- $\left({V_1 F_2 - V_2 F_2}\right) + \left({V_2 F_1 - V_1 F_1}\right) = 2 d$
But:
- $V_1 F_2 - V_2 F_2 = V_1 V_2$
- $V_2 F_1 - V_1 F_1 = V_1 V_2$
and so:
- $2 V_1 V_2 = 2 d$
By definition, the transverse axis of $K$ is $V_1 V_2$.
Hence the result.
$\blacksquare$