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)

Equation of tangent to the hyperbola
$\frac{x^2}{9}-\frac{y^2}{4}=1$ is
$y=mx+\sqrt{9m^2-4}\dots \left(i\right)$
Since, Eq. $\left(i\right)$ tagent to the circle
$\therefore$ Perpendicular distance from centre $\left(0,4,0\right)$ to the circle is equal to the radius of the circle
$\therefore \frac{|4m+0+\sqrt{9m^2-4}|}{\sqrt{1^2+m^2}}=4$
$\Rightarrow \left(4m+\sqrt{9m^2-4}\right)=4\sqrt{1+m^2}$
$\Rightarrow 16m^2+9m^2-4+8m\sqrt{9m^2-4}$
$=16\left(1+m^2\right)$
$\Rightarrow 9m^2-20$
$=-8m\sqrt{9m^2-4}$
$\Rightarrow 81m^4+400-360m^2$
$=64m^2\left(9m^2-4\right)$
$\Rightarrow 495m^4+104m^2-400=0$
$\Rightarrow m=2\sqrt{5}$
$\therefore$ From Eq. $\left(i\right)$, $y=\frac{2x}{\sqrt{3}}+\sqrt{9\times \frac{4}{5}-4}$
$\sqrt{5}y=2\times T+\sqrt{36-20}$
$\Rightarrow 2x-\sqrt{5}y+4=0$