Find the point on the parabola $2y=x^{2}$ that is closest to the point $(-4,1)$

Question

The first part of the derivative which is to the power of $-1/2$ is too small to be considered relevant. I'm not sure how to proceed from here. The answer is $(-2,2)$ but I am not sure how to get there. Thanks for the help! Edit: Sorry if this is annoying but I can't use anything but derivatives. This is a very early calculus question.

Answer 1

Minimize the distance squared $F=(x+4)^2+(y-1)^2=(x+4)^2+(\frac{x^2}{2}-1)^2=\frac{x^4}{4}+8x+17$.

$$\frac{dF}{dx}=x^3+8=0 --> x=-2$$.

$$y=2^2/2=2$$

Answer 2

Let $(x,\frac12x^2)$ be a point on the curve. The derivative at this point is $x$. And the slope of the line from $(-4,1)$ to $(x,\frac12x^2)$ is $\dfrac{\frac12x^2-1}{x+4}$.

At the nearest point, these two slopes are perpendicular: $$x\cdot\dfrac{\frac12x^2-1}{x+4} = -1$$

This gives us $\frac12x^2-x = -x - 4$, which simplifies to $x^3=-8$, with the unique solution $x=-2$.

Answer 3

let $(2a, 2a^2)$ is the point on $2y = x^2$ that has shortest point from $(-4,1).$ the slope of the tangent line at $(2a, 2a^2)$ is $2a$ this tangent must be perpendicular the line connecting $(-4,1)$ and $(2a, 2a^2).$ so the product of the slopes $$ 2a \dfrac{(2a^2 - 1)}{2a+4} = -1$$ solving this equation should give you $a = -1$ and the point $(-2, 2)$ on the parabola is the closest point to $(-4,1)$.