Question

The slope of common tangents to the hyperbolas $\frac{x^2}{9}-\frac{y^2}{16}=1$ and $\frac{y^2}{9}-\frac{x^2}{16}=1$ are:

• A. $\pm 2$
• B. $\pm 1$
• C. $\pm \sqrt{2}$
• D. None of these

Answer 1

The correct option is B $\pm 1$

Given two hyperbolas are $\frac{x^2}{9}-\frac{y^2}{16}=1$ ...(1)

and $\frac{y^2}{9}-\frac{x^2}{16}=1$ ...(2)

Equation of tangent to (1), having slope $m$ is

$y=mx\pm \sqrt{9m^2-16}$ ...(3)

Eliminating $y$ using (2) and (3), we get

$16(mx\pm \sqrt{9m^2-16})-9x^2=144$

$\Rightarrow (16m^2-9)x^2\pm 32m\sqrt{9m^2-16}+(144m^2-400)=0$

For it to be tangent, we must have $D=0$

$\therefore {\left(32m\sqrt{9m^2-16}\right)}^2=4(16m^2-9)(144m^2-400)\Rightarrow m^2=1$

$\Rightarrow m=\pm 1$