# Hyperbola problem : From a point P(1,2) pair of tangents are drawn to a hyperbola H in which one tangent to each arm of hyperbola......

Problem :

From a point P(1,2) pair of tangents are drawn to a hyperbola H in which one tangent to each arm of hyperbola. Equation of asymptotes of hyperbola H are $\sqrt{3}x -y+5=0$ and $\sqrt{3}x+y-1=0$ then find the eccentricity of hyperbola.

Solution :

Let $L_1: \sqrt{3}x -y+5=0$ $L_2: \sqrt{3}x+y-1=0$

$m_1 = \frac{-\sqrt{3}}{-1} =\sqrt{3}$ and $m_2= -\sqrt{3}$

Angle between two asymptotes is given by $tan\theta = |\frac{m_1-m_2}{1-m_1m_2}|......(1)$

After putting the values of $m_1; m_2$ in (1) we get $\theta = \frac{\pi}{3}$

We know that eccentricity e = $sec\frac{\theta}{2} = sec \frac{\pi}{6} =\frac{2}{\sqrt{3}}$

But my question what is the role of given point P(1,2) in this problem, please suggest. Thanks

Multiply both asymptotes and add constant k to make it an equation of hyperbola. Equation of Hyperbola is $$(√3x−y+5)(√3x+y−1)+k=0$$ I am not sure but what he means is that there are two tangent from point P(1,2) and four point of contact means that each arm of hyperbola has a contact tangent passing through P(1,2)
The eccentricity is $$\frac{2}{\sqrt{3}},$$ for $$\theta=\frac{\pi}{3}$$but when you can't draw tangents from $$P=(1,2)$$ the angle is $$\theta=\frac{2\pi}{3}$$and the eccentricity is $$2.$$