Question

AB is a double ordinate of the parabola y² = 4ax. Tangents drawn to parabola at A and B meets y-axis at A₁ and B₁ respectively. If the area of trapezium AA₁B₁B is equal to 12a², then angle subtended by A₁B₁ at the focus of the parabola is equal to –

• A. 2 tan⁻¹ (3)
• B. tan⁻¹ (3)
• C. 2 tan⁻¹ (2)
• D. tan⁻¹ (2)

Answer 1

Answer: (C)

2 tan⁻¹ (2)

Answer 2

Let A ŗ (at₁², 2at₁), B ŗ (at₁², –2at₁). Equation of tangents at A and B are

yt₁ = x + at₁² and yt₂ = x + at₂², respectively.

A₁ ŗ (0, at₁), B₁ ŗ (0, –at₂)

Area of trapezium AA₁ B₁B

= $\frac { 1 } { 2 }$(AB + A₁B₁) . OC

Ž 24a² = $\frac { 1 } { 2 }$. (4at₁ + 2at₁) (at₁²)

Ž t₁³ = 8 Ž t₁ = 2 Ž A₁ ŗ (0, 2a)

If ŠOSA₁ = q

Ž tan q = = 2 Ž q = tan^–1 (2).

Thus, required angle is 2tan^–1(2).