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 $y=m x+\sqrt{9 m^{2}-16}$
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
$c^{2}=a^{2} m^{2}-b^{2}$
$\Rightarrow \left(9 m^{2}-16\right)=(-16) m^{2}-(-9)$
$\Rightarrow 25 m^{2}=25$
$\Longrightarrow m=\pm 1$