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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 AA1 B1 B is equal to 12a2, then angle subtended by A1B1 at the focus of the parabola is equal to -

  1. 2 tan–1 (3)    
  2. tan–1 (3)    
  3. 2 tan–1 (2)    
  4. tan–1 (2)    

The correct answer is: 2 tan–1 (2)


    Let A ≡ (at12, 2at1), B ≡ (at12, –2at1). Equation of tangents at A and B are
    yt1 = x + at12 and yt2 = x + at22, respectively.

    A1 ≡ (0, at1), B1 ≡ (0, –at2)
    Area of trapezium AA1 B1B
    = fraction numerator 1 over denominator 2 end fraction (AB + A1B1) . OC
    rightwards double arrow 24a2 = fraction numerator 1 over denominator 2 end fraction. (4at1 + 2at1) (at12)
    rightwards double arrow t13 = 8 rightwards double arrow t1 = 2 rightwards double arrow A1 ≡ (0, 2a)
    If straight angleOSA1 = θ
    rightwards double arrow tan θ = fraction numerator 2 a over denominator a end fraction = 2 rightwards double arrow θ = tan–1 (2).
    Thus, required angle is 2tan–1(2).

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