Question

The slopes of common tangents to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ 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 $\pm 1$

For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$\

Slope form of tangent is $y=mx+c$ where condition of tangency is $c^2=a^2{m}^2-b^2...$

We have two hyperbola, $H_1:\frac{x^2}{9}-\frac{y^2}{16}=1$

and $H_2:\frac{y^2}{9}-\frac{x^2}{16}=1$

And respectively, $a_1=3,b_1=4$ and $a_2=4,b_2=3$

As the tangent is common , $c$ and $m$ remains same for both $H_1$ & $H_2$.

But in $H_2:\frac{y^2}{9}-\frac{x^2}{16}=1\Rightarrow \frac{x^2}{(-16)}-\frac{y^2}{(-9)}=1$

So, $a_1^2=9,b_1^2=4$ and $a_2^2=-16,b_2^2=-9$

Putting values in the condition of tangency,

For $H_1,c^2=a_1^2{m}^2-b_1^2\Rightarrow c^2=9m^2-16$ - Equation 1

For $H_2,c^2=a_2^2{m}^2-b_2^2\Rightarrow c^2=-16m^2-(-9)\Rightarrow c^2=-16m^2+9$ - Equation 2

From Equation 1 & 2,

$c^2=9m^2-16=-16m^2+9$

$\Rightarrow 16m^2+9m^2=16+9$

$25m^2=25$

$m^2=1$

$m=\pm 1$