Answer
$y=\pm 2 (1+b^2)^{3/2}x^2+bx$
Work Step by Step
Write formula 11.
$\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
Let us consider $f(x)=ax^2+bx$ . This is the general equation of a parabola.
Now, $f'(x)=2ax+b$ and $f''(x)=2a$
Thus,$\kappa(x)=\dfrac{|2a|}{[1+(2ax+b)^2]^{3/2}}$
or, $\kappa(0)=\dfrac{|2a|}{[1+b^2]^{3/2}}$[ $ \kappa(0)$ is the curvature at origin.]
or, $ 4=\dfrac{|2a|}{[1+b^2]^{3/2}}$
or, $a=\pm 2 (1+b^2)^{3/2}$
Therefore, the equation of parabola is $y=\pm 2 (1+b^2)^{3/2}x^2+bx$
