Question

Find the slope of common tangent if any to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Answer 1

Let tangent line of both hyperbola should be $y=x+t,y=x-t,y=-x+t,y=-x-t$

Assume, that line $y=\left(x+t\right)$ tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\frac{x^2}{a^2}-\frac{{\left(x+t\right)}^2}{b^2}=1$

$\begin{array}{c}\Rightarrow \frac{x^2}{a^2}-\frac{x^2}{b^2}-\frac{2xt}{b^2}-\frac{t^2}{b^2}-1=0\\ \Rightarrow x^2\left(b^2-a^2\right)-2a^{2}xt-a^2{t}^2-a^2{b}^2=0\end{array}$

by solving this quadratic equation we get

$\begin{array}{c}\Rightarrow t=\sqrt[+]{+a^2-b^2}ort=\sqrt[-]{-a^2-b^2},then,y=n+\sqrt{a^2-b^2}\\ \Rightarrow y=x-\sqrt{a^2-b^2},y=-x+\sqrt{a^2-b^2},y=-x-\sqrt{a^2-b^2},slope=\left(1,-1\right)\end{array}$