# Finding the points on a curve, closest to a specific point

Find the point(s) on the curve $y^3=x^2$ closest to the point $P=(0,4).$ I understand that there is a way to solve this, using the distance formula, however this turns out to seem rather complicated. I am also aware that there is a calculus method to solving this question, however am unsure as to what that method is, exactly. Any help is appreciated. Thanks :)

The square of the distance between the point $$P$$ and another point with coordinates $$(x,y)$$ is $$d^2=x^2+(y-4)^2.$$

Since the curve is defined as a set of points $$(x,y)$$ related to each other by the relation $$y^3=x^2$$, we have the square of the distance from $$P$$ to the curve equal to: $$d^2=y^3+(y-4)^2$$ Thus, the task is to find the smallest $$d$$ possible, when $$d$$ is actually a function of $$y$$. For this, we employ the derivative of $$d(y)$$ to find extremums of the function: $$d(y)=\sqrt{y^3+(y-4)^2}$$ $$d'(y)=\frac{3y^2+2(y-4)}{2\sqrt{y^3+(y-4)^2}}$$ Now, to find extremums, solve $$d'(y)=0$$, or in our case $$3y^2+2(y-4)=0$$. This quadratic equation has two solutions $$y=-2$$ and $$y=\frac{4}{3}$$. But the solution $$y=-2$$ is not acceptable, since $$y^3=x^2$$, and that would mean $$-8=x^2$$. Hence, we are left with only one solution and two pairs (do you know why?) of points:

$$(\frac{8}{\sqrt{28}},\frac{4}{3})$$ and $$(-\frac{8}{\sqrt{28}},\frac{4}{3})$$.

P.S. I did not check that obtained solution is actually a minimum. Can you validate that?

Edit: Dear reviewer, I want to edit the typo in "suare" to "square". Stupid StackExchange doesn't let me because it's less then 6 symbols!

• Thank you! Using the second derivative test I got f''(x)= 6y+2 and then subbed in the value for y, resulting in an answer of 10, thus confirming that the solution is a minimum. – user341952 May 24 '16 at 13:53

Hint:

The distance is $\sqrt{x^2+(y-4)^2}$

replace $x^2$ with $y^3$, the distance become $\sqrt{y^3+(y-4)^2}$

Let $u(y) = y^3+(y-4)^2$ find the stationary point of u(y):

$u'(y) = 3y^2+2y-8 = 0$ => $y = -2$ or $4/3$

substitute the roots into distance function and choose the minimal one.

Basic identities of derivatives