Find the equation of common tangent to the parabolas y2 = 4x and x2 =

1.	Find the equation of common tangent to the parabolas y2 = 4x and x2 = 32y.
Answer» 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 = 32 m² x + 32 ⇒ mx² – 32 m² x – 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 ⇒ 1024 m⁴ + 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.

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

Answer»

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 = 32 m² x + 32

⇒ mx² – 32 m² x – 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

⇒ 1024 m⁴ + 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.