No links please Chapter name: TANGENT AND NORMAL 30 Find the equation of the normal to the parabola - Maths - Differential Equations

Question

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C h a p t e r   n a m e :   T A N G E N T   A N D
    N O R M A L   30   F i n d   t h e
  e q u a t i o n   o f   t h e   n o r m a l
  t o   t h e   p a r a b o l a   y 2 = 4 a x
  a t   a   p o i n t   ( x 1 , y 1 )   o
n   i t   S h o w   t h a t      
          t h r e e   n o r m a l s
  c a n   b e   d r a w n   t o   a  
p a r a b o l a   f r o m   a n   e x t e r n a l
  p o i n t A n s :   y 1 x - x 1 + 2 a y - y 1 = 0

Lovina Kansal · Accepted Answer

Dear student
We have,y2=4ax    
(1)Since Px1,y1 lies on the curve (1)
Therefore,y12=4ax1     
(2)Differentiating (1) with respect to x, we get2ydydx=4a⇒dydx=2ay⇒dydxx1,y1=2ay1So, the equation of mornal at Px1,y1 isy-y1=-12ay1x-x1⇒2ay-y1=-y1(x-x1)⇒2ay-2ay1=-y1x+y1x1⇒y1x-y1x1+2ay-2ay1=0⇒y1x-x1+2ay-y1=0
Let the normlas at t1 and t2 meet at t3 on the parabola
The equation of the normal at t1 is:y+xt1=2at1at13     
(1)Equation of the chord joining t1 and t3 is:yt1+t3=2x+2at1t3      
(2)(1) and (2) represent the same line∴t1+t31=-2t1⇒t3=-t1-2t1Similarly,t3=-t2-2t2∴-t1-2t1=-t2-2t2⇒t1-t2=2t2-2t1⇒t1-t2=2t1-t2t1t2⇒t1t2=2
Regards