If all the normals drawn to the curve y= 1+3 x^2 3+x^2 at the points of intersection of y= 1+3 x^2 3+x^2 and y=1 pass through the point ( , ) , then 3 +2 =

Step-by-step solution

y= 1+3 x^2 3+x^2 To get the intersection point, let us solve aligned & y= 1+3 x^2 3+x^2 & y=1: & 1+3 x^2 3+x^2 =1 3 x^2+1=x^2+3 x= 1 & d y d x = 16 x (3+x^2 )^2 aligned Slope of the normal : m=- 1 ( d y d x ) = - (3+x^2 )^2 16 x Eq ^ n of normla is Case 1: x=1 Then, (y-1)=-1(x-1) y+x=2 Case 2: x =-1 Then y-1=1(x+1) -x+y=2 Solving eqn (i) & (ii), we get x =0 & y =2 Then ( , )=(0,2) 3 +2 =3 0+2 2=4