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)

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
$c^2=a^2{m}^2-b^2$
$\Rightarrow \left(9m^2-16\right)=(-16)m^2-(-9)$
$\Rightarrow 25m^2=25$
$⟹m=\pm 1$