Question

The slope of common tangent to the hyperbolae $\frac{x^2}{16}-\frac{y^2}{9}=1,\frac{y^2}{16}-\frac{x^2}{9}=1$ can be :

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (B), (C)

$-1$
$1$

The equation of any tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, is $y=mx\pm \sqrt{16m^2-9}$.
It is also tangent to $\frac{x^2}{9}-\frac{y^2}{16}=-1$
The equation of a tangent can be written as :
$y=mx\pm \sqrt{-9m^2+16}$
Comparing the two equations we get,
$\therefore c^2=16m^2-9=-9m^2+16$ $\Rightarrow m=\pm 1$