Closest Point on $y=x^2$ to $P(3,0)$

Question

Pico is on the parabaloa dtermined by $y=x^2$ and notices a UFO and some ancient astronauts(extraterrestrials) located at the point $P(3,0)$ in the $xy$-plane. Pico wants to get as close as possible to the UFO to take good pictures for the incredulous bureaucrats in Washington to provide compelling evidence that we are not alone in the universe. At What point on the parabola should Pico take his photos?

To my understanding, the shortest distance between any to points is the perpendicular. I.e. A $90$ degree angle is made at the intersection of the two points. If this is wrong please correct me.

So in order to find that, we set the slope of the normal line(being that the normal line is perpendicular to the tangent at a given point) at any point on $y=x^2$ equal to the slope of the line made by $P(3,0)$ and any point on $y=x^2$. $$y'=2x=m_{tan}$$$$m_{norm}=-\frac{1}{2x}=\frac{x^2-0}{x-3}=m_{line}$$ Stop my at any point. Next, we need to solve for $x$ at their intersection I believe. $$2x(x^2)=-x+3\implies 2x^3+x-3=0$$ $$(2x+3)(x-1)=0 \implies x=1,-\frac{3}{2}$$ Therefore, Pico should take photos from $P(-\frac{3}{2},\frac{9}{4})$ or $P(1,1)$ on the curve of $y=x^2$

QUESTION:

I have a couple on this. First of all is this a genuine method that works to find the shortest distance between a curve and a point and have I applied it correctly. If so, then does that mean that the two point I identified for Pico to take photos from are equal distance from $P(3,0)$? Any help or reassurance is appreciated.

Answer 1

solution using the distance formula:

according to the distance formula the distance between a function and a point is given by $$l=\sqrt{(x-x_{0})^{2}+(f(x)-y_{0})^2}$$ substituting $x_0=3$, $y_0=0$ and $f(x)=x^2$ into the equation we get $$l=\sqrt{(x-3)^{2}+(f(x)-0)^2}=\sqrt{(x-3)^{2}+(f(x))^2}$$ next in order to find the minimum distance we first take the derivitive, set in equal to zero and then solve for $x$

taking the derivative yields

$l'=\frac{\mathrm{d} }{\mathrm{d} x}{\sqrt{(x-3)^{2}+(x)^4}}=\frac{2\cdot x^3+x-3}{\sqrt{(x-3)^{2}+(x)^4}}$

setting it equal to zero

$l'=0=\frac{2\cdot x^3+x-3}{\sqrt{(x-3)^{2}+(x)^4}}$

solving for x we get $x=1$

substitute that $x$ value into $f(x)=x^2$ to the get corresponding point $y$ gives $f(1)=y=1^2=1$

so the point on $x^2$ closest to $P(3,0)$ is $(1,1)$

Answer 2

A point on the parabola has coordinates $(x,x^2)$ and its squared distance from $P(3,0)$ is given by $(x-3)^2+x^4$, that has an absolute minimum at $x=1$ (the derivative is third degree polynomial with a unique real root). It follows that the best place to take pictures is $\color{red}{(1,1)}$.

Answer 3

Hint:

$2x^3+x-3$ factorize as $(x-1)(2x^2+2x+3)$ so the unique real solution is $x=1$.

For the general question. If $P=(x_P,y_P)$ is a point and we search the point of coordinates $(x,f(x))=Q$ that has the minimum distance from $P$, we can search to minimize the square of the distance :

$D^2(x)=(x-x_P)^2+(f(x)-y_P)^2$

se we have to solve: $$ \left(D^2(x)\right)'=2(x-x_0)+2(f(x)-y_P)f'(x)=0 $$ and this gives : $$ f'(x)=-\frac{x-x_P}{f(x)+y_P} $$

and this say exactly that the slope of the tangent line at $Q$ is the opposite of the inverse of the slope of the line that passes thorough $P$ and $Q$.

But note that to be sure that this is really the minimum (and not another kind of stationary point) we have to verify it.

Answer 4

Your work is fine but $2x^3+x-3 = (x-1)(2x^2+2x +3)$.

Answer 5

A general method to solve this problem would be to use the theorem of Lagrange multiplier, even though this perpendical technique works here.

In case you are interested in the gereral technique : squared distance from $(x,y)$ to $(3,0)$ is $(3-x)^2+y^2$. Since the point has to lie on parabole $y=x^2$, Lagrange's theorem indicates that we want to minimize

$$f(x,y)=(3-x)^2+y^2+\lambda(y-x^2)$$

It has following derivatives

$$\frac{\partial f}{\partial x} = -6+2x(1-\lambda)$$ $$\frac{\partial f}{\partial y} = 2y+\lambda$$

We look for points where these derivatives are zero, which leads to

$$x=\frac{3}{1-\lambda}\qquad y = -\frac{\lambda}{2}$$

Because we have to satisfy $y=x^2$, we obtain the equation $\lambda(1-\lambda)^2=-18$, or $(\lambda+2)(\lambda^2-4\lambda+9)=0$ which has only one solution $\lambda=-2$. Therefore, you end up with only one solution $x=1,y=1$.