Finding a hyperbolic isometry that fixes the point $x = 2$ and $x = 17$

Question

I know that a Möbius transformation is hyperbolic if the trace is $> 2$ which is $a + d$. But I'm not sure of the next steps involved to arrive at the answer.

Answer 1

Hint: Consider $$f(z)=\frac{az+b}{cz+d}$$ with $ad-bc=1.$ If $c\ne 0$, then this transformation has at least one fixed point, and all fixed points will be roots of the quadratic equation $$cz^2-(a-d)z-b=0.$$

The quadratic equation you want will be of the form $$0=c(z-2)(z-17)=cz^2-(19c)+34c$$ Thus, we need $b=-34c,$ $a-d=19c$, $a+d>2$, $ad-bc=1$, and $c\ne 0$. Can you go from there?