Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqrt{a^2-b^2},y=-x+\sqrt{a^2-b^2},y=-x-\sqrt{a^2-b^2}$.

I tried to solve the two equations and find the points of intersections of the hyperbolas but these hyperbolas do not intersect each other.
Let the point of tangency be $(x_1,y_1)$ on the $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $(x_2,y_2)$ on the $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{dy}{dx}=\frac{b^2x}{a^2y}$
For $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{dy}{dx}=\frac{a^2x}{b^2y}$
$\frac{b^2x_1}{a^2y_1}=\frac{a^2x_2}{b^2y_2}$

$b^4x_1y_2=a^4x_2y_1$
I do not know how to take it further.

Hint: Implicitly differentiate each, then set the derivatives equal to one another.
