The point on the curve $x^2=2y$ which is nearest to the point $(0,5)$ is

1. Point on curve: $\left(x,\dfrac{x^2}{2}\right)$. 2. Distance squared $D=x^2+\left(\dfrac{x^2}{2}-5\right)^2$. 3. $\dfrac{dD}{dx}=2x+2\left(\dfrac{x^2}{2}-5\right)x=x(x^2-8)$. 4. $\dfrac{dD}{dx}=0 \Rightarrow x=0$ or $x^2=8$, i.e. $x=2\sqrt{2}$. 5. At $x=2\sqrt{2}$: $y=\dfrac{8}{2}=4$, giving $(2\sqrt{2},4)$, the nearest point. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.30_