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A variable tangent to the parabola x^2 = 4y intersects the lines y = x and y = -x at points P and Q respectively. The locus of the circumcentre of triangle OPQ , where O is the origin, is :

Options

  1. Ax(x^2 - y^2) + y^2 = 0
  2. B3y(x^2 - y^2) - 2x^2 = 0
  3. Cx^2(y+1) - y^3 = 0
  4. Dx^2(y-1) - y^3 = 0

Correct answer

D. x^2(y-1) - y^3 = 0

Step-by-step solution

The equation of the tangent to the parabola x^2 = 4y in slope form is y = mx - m^2 . This tangent intersects the line y = x at point P . Substituting y = x in the tangent equation: x = mx - m^2 x(m-1) = m^2 x = m^2 m-1 So, P = ( m^2 m-1 , m^2 m-1 ) . Similarly, it intersects the line y = -x at point Q . Substituting y = -x : -x = mx - m^2 x(m+1) = m^2 x = m^2 m+1 So, Q = ( m^2 m+1 , - m^2 m+1 ) . Since the lines y = x and y = -x are perpendicular, OPQ is a right-angled triangle with the right angle at the origin O

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