Question

Find the the radius of curvature of the parabola y²=4ax, at the point (at²,2at)

Answer 1

y² = 4ax

dy/dx = 2a/y

At (at², 2at), dy/dx = 2a/2at = 1/t

d²y/dx² = -2a/y² x dy/dx

At at², 2at

d²y/dx² = -2a/(4a²t²) x 1/t = -1/(2at³)

Radius of curvature R at at², 2at

= (1+(dy/dx)²)³/2/ (d²y/dx²)

= (1+1/t²)³/2/(1/(2at³))

(Negative sign may be removed)

= 2at³(t²+1)³/2/ t³

= 2a(t²+1)³/2