Question

Find the equation of tangent to the parabola y² = 12x from the point (2, 5)

Answer 1

Given equation of the parabola is y² = 12x.

Comparing this equation with y² = 4ax, we get

4a = 12

∴ a = 3

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

Since the tangent passes through the point (2, 5),

5 = `2"m" + 3/"m"`

∴ 5m = 2m² + 3

∴ 2m² – 5m + 3 = 0

∴ 2m² – 2m – 3m + 3 = 0

∴ 2m(m – 1) – 3(m – 1) = 0

∴ (m – 1)(2m – 3) = 0

∴ m = 1 or m = `3/2`

These are the slopes of the required tangents.

By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are

y – 5 = 1(x – 2) and y – 5 = `3/2("x" - 2)`

∴ y – 5 = x – 2 and 2y – 10 = 3x – 6

∴ x – y + 3 = 0 and 3x – 2y + 4 = 0