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

How do you find the points on the parabola $2x={y}^{2}$ that are closest to the point (3,0)?
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seguidora1e
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
$2s\frac{ds}{dx}=2\left(x-3\right)+2y\frac{dy}{dx}$. For minimum distance $\frac{ds}{dx}$=0, hence (x-3)+$y\frac{dy}{dx}=0$,
Or y$\frac{dy}{dx}$=3-x
Differentiating the equation ${y}^{2}$=2x with respect to x, it would be y$\frac{dy}{dx}=1$
It is thus 3-x=1, x=2, and then y=2, -2. The nearest points are (2,2) and (2,-2)

koraby2bc
An alternative starts the same as bp's solution:
Let (x,y) be any point on this parabola. The distance 's' of this point from point (3,0) is given by ${s}^{2}={\left(x-3\right)}^{2}+{y}^{2}$
Note that, since (x,y) is on the graph, we must have ${y}^{2}=2x$, so
${s}^{2}={\left(x-3\right)}^{2}+2x$
Our job now is to minimize the function:
$f\left(x\right)={\left(x-3\right)}^{2}+2x={x}^{2}-4x+9$
(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.
$f\prime \left(x\right)=2x-4$, so the critical number is 2
$f\prime \prime \left(2\right)=2$ is positive, so f(2) 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 $y=±2$
The points are: (2,2) and (2,−2)