Question

Show that the point $(\frac{a}{2}(t+\frac{1}{t}),\frac{b}{2}(t-\frac{1}{t}))$ is always on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ for $t\ne 0$ which is real.

Answer 1

By satisfying, we get the answer

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Taking LHS

$x=\frac{a}{2}(\frac{1}{t}+t)$$,y=\frac{b}{2}(-\frac{1}{t}+t)$

$\frac{1}{a^2}\left[\frac{a}{2}\right(\frac{1}{t}+t){]}^2$$-\frac{1}{b^2}\left[\frac{b}{2}\right(-\frac{1}{t}+t){]}^2$

$=\frac{1}{4}\left[2t\right][\frac{2}{t}$

$=1=$RHS