Question

The equation of the common tangent to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are

• A. $y=\pm x\pm \sqrt{b^2-a^2}$
• B. $y=\pm x\pm \sqrt{a^2-b^2}$
• C. $y=\pm x\pm (a^2-b^2)$
• D. $y=\pm x\pm \sqrt{a^2+b^2}$

Answer 1

Answer: (B)

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

Hyperbola-1 is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and

hyperbola-2 is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

We write tangent at point $(a\sec \varphi ,b\tan \varphi )$ for hyperbola 1 is

$\Rightarrow \frac{x\sec \varphi }{a}-\frac{y\tan \varphi }{b}=1$..................1

Similarly at point $(b\tan \theta ,a\sec \theta )$ for hyperbola 2 is

$\frac{y\sec \theta }{a}-\frac{x\tan \theta }{b}=1$..................2

if 1 and 2 are the common tangents they should be identical.

so $\frac{\sec \theta }{a}=-\frac{\tan \varphi }{b}\Rightarrow \sec \theta =-\frac{a\tan \varphi }{b}$...........................3

and $\frac{-\tan \theta }{b}=-\frac{\sec \varphi }{a}\Rightarrow \tan \theta =-\frac{b\sec \varphi }{a}$.........................4

we know that ${\sec }^2\theta -{\tan }^2\theta =1$

Put values from 3 and 4 in above to get,

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

$\Rightarrow {\tan }^2\varphi =\frac{b^2}{a^2-b^2},{\sec }^2\varphi =\frac{a^2}{a^2-b^2}$

So points of contact are $(\pm \frac{a^2}{\sqrt{a^2-b^2}},\pm \frac{b^2}{\sqrt{a^2-b^2}}),(\pm \frac{b^2}{\sqrt{a^2-b^2}},\pm \frac{b^2}{\sqrt{a^2-b^2}})$

length of common tangent is $\sqrt{2}\frac{(a^2+b^2)}{a^2-b^2}$

so equation of common tangent is

$\pm \frac{x}{\sqrt{a^2-b^2}}\mp \frac{y}{\sqrt{a^2-b^2}}=1$

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