Question

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

• A. $2x-\sqrt{5}y-20=0$
• B. $2x-\sqrt{5}y+4=0$
• C. $3x-4y+8=0$
• D. $4x-3y+4=0$

Answer 1

Answer: (B)

A tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ is
$y=mx+\sqrt{9m^2-4},m>0$
It is tangent to $x^2+y^2-8x=0$
$\therefore \frac{4m+\sqrt{9m^2-4}}{\sqrt{1+m^2}}=4$
$\Rightarrow 495m^4+104m^2-400=0$
$\Rightarrow m^2=\frac{4}{5}$ or $m=\frac{2}{\sqrt{5}}$
$\therefore$ the tangent is $y=\frac{2}{\sqrt{5}}m+\frac{4}{\sqrt{5}}$
$\Rightarrow 2x-\sqrt{5}y+4=0$