Question

How do you find the points on the parabola `y = 6 - x^2` that are closest to the point (0,3)?

Answer 1

Substitute the equation of the parabola into the distance formula to get the square root of a quartic to minimise. This is quadratic in `x^2`, so complete the square to find the minimum.

Explanation:

If `(x, y)` is a point on the parabola, then the distance between `(x, y)` and `(0, 3)` is:

`sqrt((x-0)^2+(y-3)^2)`

`= sqrt(x^2+(6-x^2-3)^2)`

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

`=sqrt(x^2+9-6x^2+x^4)`

`=sqrt(x^4-5x^2+9)`

`=sqrt((x^2-5/2)^2+11/4)`

This will have its minimum value when `(x^2-5/2)^2 = 0`, that is when `x = +-sqrt(5/2) = +-sqrt(10)/2`

When `x = +-sqrt(10)/2` we have `y = 6 - x^2 = 6 - 5/2 = 7/2`

So the points on the parabola at minimum distance from `(0, 3)` are:

`(+-sqrt(10)/2, 7/2)`

As a bonus, we have also calculated the minimum distance as:

`sqrt(11/4) = sqrt(11)/2`