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JEE Advanced2016MathematicsParabolaActual

Let P be the point on the parabola y 2 = 4 x which is at the shortest distance from the center S of the circle x 2 + y 2 - 4 x - 16 y + 64 = 0 . Let Q be the point on the circle dividing the line segment SP internally. Then -

Options

  1. AS P = 2 5
  2. BS Q : Q P = 5 + 1 : 2
  3. CThe x - intercept of the normal to the parabola at P is 6
  4. DThe slope of the tangent to the circle at Q is 1 2

Correct answer

A. S P = 2 5

Step-by-step solution

y 2 = 4 x Point P lies on normal to parabola passing through centre of circle y + t x = 2 t + t 3 ......(i) 8 + 2 t = 2 t + t 3 t = 2 P 4 , 4 S P = 4 - 2 2 + 4 - 8 2 S P = 2 5 S Q = 2 ⇒ P Q = 2 5 - 2 S Q Q P = 1 5 - 1 = 5 + 1 4 To find x intercept Put y = 0 in (i) ⇒ x = 2 + t 2 x = 6 ∵ Slope of common normal = - t = - 2 ∴ Slope of tangent = 1 2

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