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JEE MainMathematicsParabola

A variable normal to the parabola y^2 = 4x meets the x-axis and y-axis at points A and B respectively. The locus of the circumcentre of triangle OAB , where O is the origin, is :

Options

  1. Ay^2 = 2x^2(x-1)
  2. By^2 = x^2(3x-2)
  3. C2y^2 + x = 0
  4. Dy^2 = x^2(x-2)

Correct answer

A. y^2 = 2x^2(x-1)

Step-by-step solution

The equation of the normal to the parabola y^2 = 4x in slope form is y = mx - 2m - m^3 . This normal meets the x-axis ( y = 0 ) at point A . 0 = mx - 2m - m^3 x = 2 + m^2 (for m 0 ) So, A = (2+m^2, 0) . It meets the y-axis ( x = 0 ) at point B . y = -2m - m^3 So, B = (0, -2m-m^3) . Since the coordinate axes are perpendicular, OAB is a right-angled triangle with the right angle at the origin O . The circumcentre (h, k) of a right-angled triangle is the midpoint of its hypotenuse AB . h = 2+m^2 + 0 2 2h = 2 + m^2 m^2

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