Two tangents to the hyperbola x^2/a^2 - y^2/b^2 = 1, make angles θ1 , θ2 , with the transverse axis.

Question

Two tangents to the hyperbola \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) make angles θ₁ , θ₂ , with the transverse axis. Find the locus of their point of intersection if tan θ₁ + tan θ₂ = k.

x²/a² - y²/b² = 1

Answer 1

Given equation of the hyperbola is\(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Let θ₁ and θ₂ be the inclinations. m₁ = tan θ₁ , m₂ = tan θ₂

Let P(x₁ , y₁) be a point on the hyperbola 

Equation of a tangent with slope ‘m’ to the hyperbola \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) is

y = mx ± \(\sqrt{a^2m^2- b_2}\)

This tangent passes through P(x₁ , y₁).

This is a quadratic equation in ‘m’. 

It has two roots say m₁ and m₂, which are the slopes of two tangents drawn from P.

∴ m₁+m₂ = \(\frac {2x_1y_1}{x^2_1-a^2}\)

Since tan θ₁ + tan θ₂ = k,

\(\frac {2x_1y_1}{x^2_1-a^2}\)= k

∴ P(x₁ , y₁) moves on the curve whose equation is k(x² – a²) = 2xy.