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

##### 1 Answer
Aug 16, 2015

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 $\left(x , y\right)$ is a point on the parabola, then the distance between $\left(x , y\right)$ and $\left(0 , 3\right)$ is:

$\sqrt{{\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2}}$

$= \sqrt{{x}^{2} + {\left(6 - {x}^{2} - 3\right)}^{2}}$

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

$= \sqrt{{x}^{2} + 9 - 6 {x}^{2} + {x}^{4}}$

$= \sqrt{{x}^{4} - 5 {x}^{2} + 9}$

$= \sqrt{{\left({x}^{2} - \frac{5}{2}\right)}^{2} + \frac{11}{4}}$

This will have its minimum value when ${\left({x}^{2} - \frac{5}{2}\right)}^{2} = 0$, that is when $x = \pm \sqrt{\frac{5}{2}} = \pm \frac{\sqrt{10}}{2}$

When $x = \pm \frac{\sqrt{10}}{2}$ we have $y = 6 - {x}^{2} = 6 - \frac{5}{2} = \frac{7}{2}$

So the points on the parabola at minimum distance from $\left(0 , 3\right)$ are:

$\left(\pm \frac{\sqrt{10}}{2} , \frac{7}{2}\right)$

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

$\sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$