Question

Find the equation and the length of the common tangents 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

Tangent at $(a\sec \varphi ,b\tan \varphi )$ on the 1 st hyperbola is $\frac{x}{a}\sec \varphi -\frac{y}{b}\tan \varphi =1$

Similarly tangent at any point $(b\tan \theta ,a\sec \theta )$ on 2nd hyperbola is $\frac{y}{a}\sec \theta -\frac{x}{b}\tan \theta =1$

If (1) and (2) are common tangents then they should be identical. Comparing the co-effecients of x and y

$\Rightarrow \frac{\sec \theta }{a}=-\frac{\tan \varphi }{b}$ and $-\frac{\tan \theta }{b}=\frac{\sec \varphi }{a}or\sec \theta =-\frac{a}{b}\tan \varphi \because {\sec }^2\theta -{\tan }^2\theta =1\Rightarrow \frac{b^2}{a^2}{\tan }^2\varphi -\frac{b^2}{a^2}{\sec }^2\varphi =1$

$\frac{a^2}{b^2}{\tan }^2\varphi -\frac{b^2}{a^2}(1+{\tan }^2\varphi )=1or(\frac{a^2}{b^2}-\frac{b^2}{a^2}){\tan }^2\varphi =1+\frac{b^2}{a^2}{\tan }^2\varphi =\frac{b^2}{a^2-b^2}and{\sec }^2\varphi =1+{\tan }^2\varphi =\frac{a^2}{a^2-b^2}$

Hence the point of contact are

$\{\pm \frac{a^2}{\sqrt{(a^2-b^2)}},+\frac{b^2}{\sqrt{(a^2-b^2)}}\}and\{\pm \frac{b^2}{\sqrt{(a^2-b^2)}},\pm \frac{a^2}{\sqrt{(a^2-b^2)}}\}$

Length of common tangent i.e., the distance between the above points is $\sqrt{2}\frac{(a^2+b^2)}{\sqrt{(a^2-b^2)}}$ and equation of common tangent on putting the values of $\sec \varphi and\tan \varphi$ in (1) is $\pm \frac{x}{\sqrt{(a^2-b^2)}}\mp \frac{y}{\sqrt{(a^2-b^2)}}=1orx\mp y=\pm \sqrt{(a^2-b^2)}$