## Calculus: Early Transcendentals 8th Edition

$y=2x^2$
In order to find the equation of the parabola we need to find the curvature ,for this we will have to use formula 11, such that $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ Consider $f(x)=ax^2+bx$ [ General equation of a parabola] $f'(x)=2ax+b$ and $f''(x)=2a$ $\kappa(x)=\dfrac{|2a|}{[1+(2ax+b)^2]^{3/2}}$ As we are given that the $\kappa(0)$ is the curvature at origin. Therefore, $\kappa(0)=\dfrac{|2a|}{[1+b^2]^{3/2}}$ $\implies 4=\dfrac{|2a|}{[1+b^2]^{3/2}}$ or, $a=\pm 2 (1+b^2)^{3/2}$ Hence, the equation of parabola will be: $y=\pm 2 (1+b^2)^{3/2}x^2+bx$ If we take the parabola to have the vertex at the origin, then $b=0$ and $y=2x^2$