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Question

The equations of the normals at the ends of the latus rectum of the parabola y2 = 4ax are given by

  1. x2 – y2 – 6ax + 9a2 = 0    
  2. x2 – y2 – 6ax – 6ay + 9a2 = 0    
  3. x2 – y2 – 6ay + 9a2= 0    
  4. none of these    

hintHint:

Find the end points of latus rectum of given parabola ,Also find the slope of normal at that point to form two equations.

The correct answer is: x2 – y2 – 6ax + 9a2 = 0


    The co-ordinates of end points of latus rectum is (a, 2a) and (a,-2a) .
    Then slope of tangent of y2 = 4ax is  2 y fraction numerator d y over denominator d x end fraction space equals space 4 a space rightwards double arrow fraction numerator d y over denominator d x end fraction space equals space fraction numerator 2 a over denominator y end fraction space left parenthesis h e r e comma space y space equals plus-or-minus 2 a right parenthesis
    Slope of tangent at (a, 2a) and (a,-2a) are 1 and -1 respectively.
    As, Product of Slopes of perpendicular lines is -1.
    Then Slope of Normal at (a, 2a) and (a,-2a) are fraction numerator negative 1 over denominator 1 end fraction space equals space minus 1 and fraction numerator negative 1 over denominator negative 1 end fraction space equals 1  respectively.
    Equations of normal are y - 2a = -1(x-a) => y+ x -3a = 0 and y + 2a = 1(x-a) => x-y -3a =0
    Then equation of pair of lines is  (y+ x -3a)( x-y -3a ) = 0
    The equations of the normals at the ends of the latus rectum of the parabola y2 = 4ax are given by
    x- y2- 6ax + 9a2 = 0 .

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