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

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= (x-3)^2 +y^2`. Differentiate both sides w.r.t x

`2s(ds)/dx= 2(x-3) + 2ydy/dx`. For minimum distance `(ds)/dx` =0, hence (x-3)+`ydy/dx=0`, Or y`dy/dx`= 3-x

Differentiating the equation `y^2`=2x with respect to x, it would be y`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)

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= (x-3)^2 +y^2`.

Note that, since `(x,y)` is on the graph, we must have `y^2=2x`, so

`s^2= (x-3)^2 +2x`

Our job now is to minimize the function:

`f(x) = (x-3)^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'(x) = 2x-4`, so the critical number is `2`

`f''(2) = 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(2)=y^2` has solutions `y=+-2`

The points are: `(2,2)` and `(2,-2)`