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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

Sachin
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2 Answers2

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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)

See the definition "A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows. Each branch of the hyperbola has two arms which become straighter (lower curvature) further out from the center of the hyperbola. Diagonally opposite arms, one from each branch, tend in the limit to a common line, called the asymptote of those two arms. So there are two asymptotes, whose intersection is at the center of symmetry of the hyperbola."

Samar Imam Zaidi
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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 $2.$

Jan-Magnus Økland
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