Question

Answer the following:

The slopes of the tangents drawn from P to the parabola y² = 4ax are m₁ and m₂, show that  m₁ − m₂ = k, where k is a constant.

Answer 1

Let P(x₁, y₁) be any point on the parabola y² = 4ax

Equation of tangent to the parabola y² = 4ax having slope m is y = `"m"x + "a"/"m"`

This tangent passes through P(x₁, y₁)

∴ y₁ = `"m"x_1 + "a"/"m"`

∴ my₁ = m²x₁ + a

∴ m²x₁ – my₁ + a = 0

This is a quadratic equation in ‘m’.

The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents drawn from P.

∴ m₁ + m₂ = `y_1/x_1`, m₁·m₂ = `"a"/x_1`

(m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂

= `(y_1/x_1)^2 - (4"a")/x_1`

= `(y_1^2 - 4"a"x_1)/x_1^2`

∴ m₁ – m₂ = `sqrt((y_1^2 - 4"a"x_1)/x_1^2)`

Since (x₁, y₁) and a are constants, m₁ − m₂ is a constant.

∴ m₁ – m₂ = k, where k is constant.