Hyperbola Caustic

The right-hand section of the hyperbola can be parametrized as $(\sec t, \tan t)$ for $-\tfrac12\pi < t < \tfrac12\pi$ . The tangent to the hyperbola at the point $P\;(\sec t,\tan t)$ is $\mathrm{cosec}\,t = \tan\alpha$ , and so the reflection of a horizontal light ray incident at $P$ has gradient $\tan2\alpha = -2\sec t \tan t$ . Thus the equation of the reflected light ray at $P$ is $F(x,y,t) \; = \; 2\sec t \tan t x + y - \tan t + 2\sec^2t \tan t \; = \; 0$ We need to obtain the envelope of this family of curves, and we do this by solving the simultaneous equations $F(x,y,t) \; = \; 0 \hspace{2cm} \frac{\partial F}{\partial t}(x,y,t) \; = \; 0$ for $x$ and $y$ . If we do these, we obtain the parametric equation of the catacaustic $x \; = \; \tfrac32\sec t \hspace{2cm} y \; = \; -\tan^3t$ Thus the Cartesian equation of the catacaustic is $x \; = \; \tfrac32\sqrt{1 + |y|^{\frac23}}$ Of course, the right-hand section of the hyperbola has Cartesian equation $x \; = \; \sqrt{1 + y^2}$ and we can solve these two equations simultaneously to see that the catacaustic meets the hyperbola when $y = \pm\alpha = \pm \tfrac{1}{2} \sqrt{\tfrac{1}{2} (19 + 9 \sqrt{6})}$ . Thus we need to calculate $A \; = \; \int_{-\alpha}^\alpha \big(\tfrac32\sqrt{1 + |y|^{\frac23}} -\sqrt{1 + y^2}\big)\,dy$ which can be evaluated exactly: $A \; = \; \frac{1}{16} \left[ 3 \sqrt{390 + 160 \sqrt{6}} - 16\mathrm{cosech}^{-1}\big(\tfrac{2}{5} \sqrt{\tfrac25 (-19 + 9 \sqrt{6})}\big) - 18 \sinh^{-1}\big(\sqrt{\tfrac{1}{2} (1 + \sqrt{6})}\big) \right]$ or numerically, so $A = \boxed{2.464404358}$ .