Question

Find the equation of tangent to the parabola y² = 36x from the point (2, 9)

Answer 1

Given the equation of the parabola is y² = 36x.

Comparing this equation with y² = 4ax, we get

4a = 36

∴ a = 9

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

Since the tangent passes through the point (2, 9),

9 = `2"m" + 9/"m"`    

∴ 9m = 2m² + 9

∴ 2m² – 9m + 9 = 0

∴ 2m² – 6m – 3m + 9 = 0

∴ 2m(m – 3) – 3(m – 3) = 0

∴ (m – 3)(2m – 3) = 0

∴ m = 3 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 – 9 = 3(x – 2) and y – 9 = `3/2("x" - 2)`

∴ y – 9 = 3x – 6 and 2y – 18 = 3x – 6

∴ 3x – y + 3 = 0 and 3x – 2y + 12 = 0