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Question

The equation of the normal having slope m of the parabola y2 = x + a is

  1. y = mx – am – am3    
  2. y = mx – 2am – am3    
  3. 4y = 4mx + 4am – 2m – m3    
  4. 4y = 4mx + 2am –am3    

hintHint:

equation of normal:
y= mX-2am-am^3

The correct answer is: 4y = 4mx + 4am – 2m – m3



    4y = 4mx + 4am – 2m – m3
    replace x+a by X
    the equation becomes
    y^2 = 4(1/4)(X)
    therfore, equation of normal:
    y= mX-2am-am^3
    y= m(x+a) -2(1/4)m- (1/4)m^3
    on solving, we get
    4y = 4mx + 4am – 2m – m3

    parametric form of slope of parabola = y+tx = 2at + at^3

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