Question

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. if the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is ______.

Options

• 3(x + y) + 4 = 0

• 8(2x + y) + 3 = 0

• 4(x + y) + 3 = 0

• x + 2y + 3 = 0

Answer 1

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. if the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is 4(x + y) + 3 = 0.

Explanation:

As origin is the only common point to x-axis and y-axis, so, origin is the common vertex

Let the equation of two parabolas be y² = 4ax and x² = 4by

Now latus rectum of both parabolas = 3

∴ 4a = 4b 

⇒ a = b = `3/4`

∴ Two parabolas are y² = 3x and x² = 3y

Suppose y = mx + c is the common tangent.

∴ y² = 3x

⇒ (mx + c)² = 3x

⇒ m²x² + (2mc – 3)x + c² = 0

As, the tangent touches at one point only

So b² – 4ac = 0

⇒ (2mc – 3)² – 4m²c² = 0

⇒ 4m²c² + 9 – 12mc – 4m²c² = 0

⇒ c = `9/(12m)` = `3/(4m)`  ...(i)

∴ x² = 3y

⇒ x² = 3(mx + c)

⇒ x² – 3mx – 3c = 0

Again, b² – 4ac = 0,

⇒ 9m² – 4(1)(–3c) = 0

⇒ 9m² = –12c  ...(ii)

Form (i) and (ii)

m² = `(-4c)/3` = `(-4)/3(3/(4m))` 

⇒ m³ = –1

⇒ m = –1

⇒ c = `(-3)/4`

Hence, y = mx + c = `-x - 3/4`

⇒ 4(x + y) + 3 = 0