Mathematical Reflections 2012, Issue 4 - Problem U237

Problem:
Let $\mathcal{H}$ be a hyperbola with foci $A$ and $B$ and center $O$. Let $P$ be an arbitrary point
on $\mathcal{H}$ and let the tangent of $\mathcal{H}$ through $P$ cut its asymptotes at $M$ and $N$. Prove that $PA + PB = OM + ON$.

Proposed by Luis Gonzalez.

Solution:

Let $\mathcal{H}: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, so that $O=(0,0)$. By symmetry, suppose that $P$ is on the first quadrant. Then $P=\left(k,\dfrac{b}{a}\sqrt{k^2-a^2}\right)$, where $k \geq a$ is an arbitrary real number. The equation of the tangent of $\mathcal{H}$ at $P$ is $\dfrac{kx}{a^2}-\dfrac{\dfrac{b}{a}\sqrt{k^2-a^2}y}{b^2}=1$, so intersecting this line with the asymptotes $y=\pm \dfrac{b}{a}x$ we get the points $$M=\left(k+\sqrt{k^2-a^2},\dfrac{b}{a}(k+\sqrt{k^2-a^2})\right), \qquad N=\left(k-\sqrt{k^2-a^2},-\dfrac{b}{a}(k-\sqrt{k^2-a^2})\right).$$ Then
$$OM+ON=\dfrac{2k}{a}\sqrt{a^2+b^2}=2ke,$$ where $e=\sqrt{a^2+b^2}/a$ is the eccentricity of the hyperbola. Since the foci have coordinates $A=(-c,0), B=(c,0)$ where $c=\sqrt{a^2+b^2}$, we have
$$\begin{array}{lcl}PA+PB & = &\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}+\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)} \\ & = & \dfrac{4kc}{\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}-\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}}\\ &=&\dfrac{2kc}{a}\\&=&2ke \end{array}$$
and the statement follows.