Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an orthogonal grid: what am I doing wrong?

Question

Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an orthogonal grid.

My idea is to find the intersection points of the hyperbolae, and then take a look at the $y_1' \cdot y_2'$ at those points. It should be $-1$.

From $\begin{cases} |y|=\sqrt{x^2-a} \\ y=\frac{b}{x} \end{cases}$ we get $\sqrt{x^2-a}=\pm \frac{b}{x}$.

My derivatives are $y_1'=\frac{1}{\sqrt{x^2-a}}$, $y_2'= \frac{b}{x^2}$.

$y_1'\cdot y_2' = \pm \frac{x}{b} \cdot \frac{b}{x^2} = \pm \frac{1}{x} \neq -1$.

I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray.

Thank you.

Answer 1

Your curves are level lines of the functions $$F(x,y):=x^2-y^2,\qquad G(x,y):=xy\ .$$ Given a point ${\bf z}_0=(x_0,y_0)\ne{\bf 0}$ the level line of $F$ through ${\bf z}_0$ is orthogonal to $\nabla F({\bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${\bf z}_0$ is orthogonal to $\nabla G({\bf z}_0)=(y_0,x_0)$. As $\nabla F({\bf z}_0)\cdot\nabla F({\bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.

Answer 2

These are the real and imaginary parts of $z\mapsto z^2$. Then use the Cauchy-Riemann equations.