Question

Find the equation of common tangent to the parabola y² = 4x and x² = 32y

Answer 1

Given equation of the parabola is y² = 4x.

Comparing this equation with y² = 4ax, we get

4a = 4

∴ a = 1

Let the equation of common tangent be

y = `"mx" + 1/"m"`   ...(i)

Substituting y = `"mx" + 1/"m"` in x² = 32y, we get

x² = `32("mx" + 1/"m")`

= `32  "mx" + 32/"m"`

∴ mx² = 32m²x + 32

∴ mx² – 32m²x – 32 = 0 …(ii)

Line (i) touches the parabola x² = 32y.

∴ The quadratic equation (ii) in x has equal roots.

∴ Discriminant = 0

∴ (– 32m²)² – 4(m) (– 32) = 0

∴ 1024m⁴ + 128m = 0

∴ 128m (8m³ + 1) = 0

∴ 8m³ + 1 = 0   ...[∵ m ≠ 0]

∴ m³ = `-1/8`

∴ m = `-1/2`

Substituting m = `-1/2` in (i), we get

y = `-1/2"x" + 1/((-1/2)`

∴ y = `-1/2 "x" - 2`

∴ x + 2y + 4 = 0, which is the equation of the common tangent.