Question

The slope of common tangents of hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ and $\frac{y^2}{9}-\frac{x^2}{16}=1$ is

• A. $2,-2$
• B. $1,-1$
• C. $1,2$
• D. $-1,-2$

Answer 1

Answer: (B)

$1,-1$

The equation of tangent to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ is $y=mx+\sqrt{9m^2-16}$
If it also be tangent to $\frac{y^2}{9}-\frac{x^2}{16}=1$ ie, $\frac{x^2}{(-16)}-\frac{y^2}{(-9)}=1$. Then, we get
$c^2=a^2{m}^2-b^2$
$\Rightarrow (9m^2-16)=(-16)m^2-(-9)$
$\Rightarrow 25m^2=25$
$\Rightarrow m=\pm 1$