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BITSAT2022MathematicsParabolaActual

A normal is drawn at the point P to the parabola y^2=8 x , which is inclined at 60^ with the straight line y=8 . Then the point P lies on the straight line

Options

  1. A2 x+y-12-4 3 =0
  2. B2 x-y-12+4 3 =0
  3. C2 x-y-12-4 3 =0
  4. DNone of these

Correct answer

C. 2 x-y-12-4 3 =0

Step-by-step solution

For the parabola y^2=4 a x , the equation of normal at P (a m^2,-2 a m ) is y=m x-2 a m-a m^3 . Here, m= 60^ = 3 P (a( 3 )^2,-2 a( 3 ) ) (6,-4 3 ) ( a=2) Thus, P satisfies 2 x-y-12-4 3 =0

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