Question

Answer the following:

Two tangents to the hyperbola `x^2/"a"^2 - 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

Answer 1

Given equation of the hyperbola is `x^2/"a"^2 - 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

`x^2/"a"^2 - y^2/"b"^2` = 1 is y = `"m"x ± sqrt("a"^2"m"^2 - "b"^2)`

This tangent passes through P(x₁, y₁).

∴ y₁ = `"m"x_1 ± sqrt("a"^2"m"^2 - "b"^2)`

∴ (y₁ – mx₁)² = a²m² – b²

∴ (`"x"_1^2` – a²)m² – 2x₁y₁m + (`"y"_1^2` + b²) = 0 …(i)

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 = `(2"x"_1"y"_1)/("x"_1^2-"a"^2)`

Since tan θ₁ + tan θ₂ = k,

`(2"x"_1"y"_1)/("x"_1^2-"a"^2)` = k

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