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General
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Question

PA and PB are the tangents drawn to y2 = 4x from point P. These tangents meet the y-axis at the points A1 and B1 respectively. If the area of triangle PA1 B1 is 2 sq. units, then locus of ‘p’ is -

  1. a2 (y2 + 2x)x2 = 8    
  2. a2 (y2 + 4x)x2 = 16    
  3. a2 (y2 – 4x)x2 = 8    
  4. a2 (y2 – 4x)x2 = 16    

The correct answer is: a2 (y2 – 4x)x2 = 16


    Let A ≡ (t12, 2t1), B ≡ (t22, 2t2), then P ≡ (t1t2, t1 + t2).
    Also equations of PA and PB are yt1 = x + at12, yt2 = x + at22.
    Thus, A1 ≡ (0, at1), B1 ≡ (0, at2) .
    Now, area of triangle PA1B1 = fraction numerator 1 over denominator 2 end fraction. |A1B1| |t1 t2|
    = fraction numerator 1 over denominator 2 end fractiona |t1 – t2| |t1 t2| = 2(given)
    rightwards double arrow a2 (t1 – t2)2 (t1t2)2 = 16
    rightwards double arrow a2 [(t1 + t2)2 – 4t1t2] (t1t2)2 = 16
    Thus locus of ‘P’ is a2 (y2 – 4x) x2 = 16.

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