Question

Find the point on the parabola y = (x-3)^2 that is closest to the origin...

Answer 1

distance from origin
\[d = \sqrt{x^2 +y^2}\]
minimize distance by setting derivative equal to 0
\[\frac{2x +2y \frac{dy}{dx}}{2 \sqrt{x^2+y^2}} = 0\]
\[\rightarrow \frac{dy}{dx} = -\frac{x}{y}\]
\[2(x-3) = -\frac{x}{(x-3)^2}\]
\[2x^3 -18x^2 +55x -54=0\]
solve for x