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Through the vertex O of the parabola y 2 = 4 a x chords OP and OQ are drawn such that OP is perpendicular to OQ . For each and every possible chords OP and OQ , the chord PQ cuts the axis of parabola at a fixed point, whose coordinates are

Options

  1. A( 2 a ,   0 )
  2. B( 6 a ,   0 )
  3. C( 4 a ,   0 )
  4. D( 8 a ,   0 )

Correct answer

C. ( 4 a ,   0 )

Step-by-step solution

Given parabola is y 2 = 4 a x . Any point on the above parabola can be assumed as ( a t 2 ,   2 a t ) . Let us assume that the coordinates of points P and Q are P a t 1 2 ,  2 a t 1 and Q a t 2 2 ,  2 a t 2 . Therefore, slope of O P = 2 a t 1 - 0 a t 1 2 - 0 = 2 t 1 and slope of O Q = 2 a t 2 - 0 a t 2 2 - 0 = 2 t 2 Since, O P ⊥ O Q , hence 2 t 1 × 2 t 2 = - 1 ⇒ t 1 t 2 = - 4 Also, the equation of the chord joining P and Q is, y - 2 a t 1 = 2 a t 2 - 2 a t 1 a t 2 2 - a t 1 2 x - a t

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