# How do you find the points on the parabola 2x = y^2 that are closest to the point (3,0)?

May 13, 2015

Points are (2.2) and (2, -2)

Let there be any point (x,y) on this parabola. The distance 's' of this point from point (3,0) is given by ${s}^{2} = {\left(x - 3\right)}^{2} + {y}^{2}$. Differentiate both sides w.r.t x

$2 s \frac{\mathrm{ds}}{\mathrm{dx}} = 2 \left(x - 3\right) + 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$. For minimum distance $\frac{\mathrm{ds}}{\mathrm{dx}}$ =0, hence (x-3)+$y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, Or y$\frac{\mathrm{dy}}{\mathrm{dx}}$= 3-x

Differentiating the equation ${y}^{2}$=2x with respect to x, it would be y$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

It is thus 3-x=1, x=2, and then y=2, -2. The nearest points are (2,2) and (2,-2)

May 14, 2015

An alternative starts the same as bp's solution:

Let $\left(x , y\right)$ be any point on this parabola. The distance 's' of this point from point $\left(3 , 0\right)$ is given by ${s}^{2} = {\left(x - 3\right)}^{2} + {y}^{2}$.

Note that, since $\left(x , y\right)$ is on the graph, we must have ${y}^{2} = 2 x$, so

${s}^{2} = {\left(x - 3\right)}^{2} + 2 x$

Our job now is to minimize the function:

$f \left(x\right) = {\left(x - 3\right)}^{2} + 2 x = {x}^{2} - 4 x + 9$

(It should be clear that we can minimize the distance by minimizing the square of the distance.)

To minimize, differentiate, find and test critical numbers.

$f ' \left(x\right) = 2 x - 4$, so the critical number is $2$

$f ' ' \left(2\right) = 2$ is positive, so $f \left(2\right)$ is a minimum.

The question asks for points on the graph, so we finish by finding points on the graph at which $x = 2$

$2 \left(2\right) = {y}^{2}$ has solutions $y = \pm 2$

The points are: $\left(2 , 2\right)$ and $\left(2 , - 2\right)$