Originally Posted by

**Jacobpm64** Find the coordinates of the point on the parabola $\displaystyle y = x^2$ which is the closest to the point (3,0).

I'm just wondering if I did this correctly:

a point on the curve would be $\displaystyle (x,x^2)$

So, i found the distance between $\displaystyle (3,0) $ and $\displaystyle (x,x^2)$.

$\displaystyle d = \sqrt{(0-x^2)^2+(3-x)^2}$

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

Then, I took the derivative to minimize the distance.

$\displaystyle d' = \frac{1}{2} (x^4 + 9 - 6x + x^2)^{-\frac{1}{2}} * (4x^3 - 6 + 2x) $

Set it equal to zero to find where the minimum occurs.

$\displaystyle \frac{4x^3 - 6 + 2x}{2 \sqrt{x^4 + 9 - 6x + x^2}} = 0$

$\displaystyle 4x^3 - 6 + 2x = 0$

$\displaystyle x = 1$

Plug it back in to get the y-value

$\displaystyle y(1) = 1^2 = 1$

The point on $\displaystyle y = x^2$ which is closest to the point $\displaystyle (3,0)$ is $\displaystyle (1,1)$.

Is this correct?