Question

The chord of the parabola y = −p²x² + 5px − 4 touches the curve y = 1/(1− x) at the point x = 2 and is bisected by that point. Find the number of the values of ‘p’.

Answer 1

Given

y = −p²x² + 5 px − 4 (1)

y = 1/(1− x)    (2)

Chord touches curve (2) at x = 2 which gives y = −1.

Let (x₁, y₁) and (x₂, y₂) are ends of chord.

Touching point is middle point ((x₁ + x₂)/2, (y₁ + y₂)/2)

(x₁ + x₂)/2 = 2; x₁ + x₂ = 4

and y₁ + y₂ = -2

(x₁, y₁) and (x₂, y₂) satisfy the curve.

Therefore,

y₁ = −p²x²₁ + 5px₁ − 4 (3)

and y₂ = −p²x²₂ + 5px₂ − 4 (4)

Subtracting Eq. (4) from Eq. (3), we get

y₁ − y₂ = − p²x²₁ + 5px₁ − 4 + p²x²₂ − 5px₂ + 4

⇒ y₁ − y₂ = −p²(x²₁ − x²₂) + 5p(x₁ − x₂)

⇒ (y₁ - y₂)/(x₁ - x₂) = − p²(x₁+ x₂) + 5p = −4p² + 5p (5)

Again, (dy/dx)_{at x=2}  = 1/(1 - x)² = 1/1

Therefore, from Eq. (5), we get

1 = −4p² + 5p

⇒ 4p² − 5p + 1 = 0

⇒ p = 1, 1/4