Question

find the radius of curvature at any point p(x,y) one the parabola y^2=4ax​

Answer 1

Given : parabola y^2=4ax​

To Find :  radius of curvature at any point p(x,y)

Solution:

y² = 4ax

=> 2y dy/dx  = 4a

=> dy/dx  = 4a/2y

=> dy/dx = 2a/y

dy/dx  = 2a/y

=> d²y/dx²   =  (-2a/y²) (dy/dx)

=> d²y/dx²   =   (-2a/y²)(2a/y)

=>  d²y/dx²   =   -4a²/y³

R = radius of curvature

R =  (  1  + (dy/dx)²)^(3/2) / (d²y/dx²)

=> R =  ( 1  +  (2a/y)²)^(3/2) / ( -4a²/y³)

=> R =  ( y² +  4a²)^(3/2) / ( -4a² )

=> R = - ( y² +  4a²)^(3/2) / 4a²

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