Question

Find the equation, the length, and the common tangent to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.

Answer 1

Let $y=mx+c$ be the common hyperbola

We know that condition for $y=mx+c$ to be tangent to a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $a^2{m}^2-b^2=c^2$

$\therefore$ for first hyperbola $c^2=a^2{m}^2-b^2$ ----equation 1

and for second hyperbola, $c^2=a^2-b^2{m}^2$ ----equation2

Equating the 2 equations, we get $a^2{m}^2-b^2=a^2-m^2{b}^2$

$\Rightarrow (a^2+b^2)m^2=a^2+b^2$

$a^2+b^2\ne 1$

$\Rightarrow m=\pm 1\&c^2=a^2-b^2$

Putting it in the equation of $y=mx+c$, we get the equation of tangent as

$y=\pm x\pm \sqrt{a^2-b^2}$

To find the length of tangents-

We can rewrite the tangents to $\pm \frac{y}{\sqrt{a^2-b^2}}\mp \frac{x}{\sqrt{a^2-b^2}}=1$

We know that tangent to the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $(x_1,y_1)$ is given by $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$

On comparing with first hyperbola, we get $x_1=\pm \frac{a^2}{\sqrt{a^2-b^2}}\&y_1=\pm \frac{b^2}{\sqrt{a^2-b^2}}$

and for second hyperbola we get the corresponding points $x_2=\pm \frac{b^2}{\sqrt{a^2-b^2}}\&y_2=\pm \frac{a^2}{\sqrt{a^2-b^2}}$

Using distance formula, we get length of the common tangent $=(a^2+b^2)\sqrt{\frac{2}{a^2-b^2}}$

and the equation of common tangent is $y=\pm x\pm \sqrt{a^2-b^2}$