# Shortest distance between parabola and point

Find the shortest distance between the parabola defined by $y^2 = 2x$ and a point $E:= (1.5, 0)$.

I can't use the distance formula because I'm missing a set of points $(x, y)$ to plug into. So, instead, I have a normal that passes through the point $E$ from the parabola. Which is the definition of the shortest distance to a point.

$$y - y_1 = m(x - x_1)$$

The slope of the normal is $\frac{1}{y_1}$ by using implicit differentiation and that's where I'm stuck, because I plug the point E into it and I get

$$y_1^2=x_1-1.5$$

How do I prove the shortest distance is $\sqrt{2}$?

• Nitpick: how do you define the "distance" between an equation and a point? (You really mean the distance between the parabola and the point.) – Andreas Rejbrand Jun 11 '18 at 11:05

For a point $(x,y)$ on the parabola we have that

$$d^2=(y-0)^2+\left(x-\frac32\right)^2=y^2+x^2-3x+\frac94=x^2-x+\frac94$$

and the minimum is attained for $x=\frac12$ thus at the point $(\frac12,1)$ and therefore

$$d=\sqrt{1^2+1^2}=\sqrt 2$$

• Note that there are two points – $(\frac 12, 1)$ and $(\frac 12, -1)$ – where the same minimum distance is attained, see the image in José's answer. – Paŭlo Ebermann Jun 11 '18 at 21:09
• @PaŭloEbermann Yes of course by symmetry, indeed the distance for y positive or negative is the same $$d=\sqrt{(y-0)^2+\left(x-\frac32\right)^2}=\sqrt{(-y-0)^2+\left(x-\frac32\right)^2}$$ – user Jun 11 '18 at 21:13

Given a point $P=\left(\frac{y^2}2,y\right)$ of your parabola, consider the line segment joining $P$ to $C=\left(\frac32,0\right)$. The slope of this line segment is $\frac{2y}{y^2-3}$. And the slope of the tangent to the parabola at $P$ is $\frac1y$. Since two lines are orthogonal if and only if one of them is horizontal and the other one is vertical or when the product of their slopes is $-1$, these lines are orthogonal if and only if $y=0$ or $\frac2{y^2-3}=-1$, which means that $y=0$ or that $y=\pm1$. Forget $0$: that's a local maximum. So, the distance from the parabola to $C$ is$$\left\|\left(\frac12,1\right)-\left(\frac32,0\right)\right\|=\sqrt2.$$

• +1 for picture. – Matthew Leingang Jun 11 '18 at 12:28
• @MatthewLeingang Also for the answer :-) – Sebastiano Oct 6 '20 at 22:38

Any point of the parabola is of the type $(x,Y)$ with $y^{2}=2x$. The distance between $(x,y)$ and $(1.5,0)$ is $\sqrt (x-1.5)^{2}+y^{2}=(x-1.5)^{2}+2x$. You have to find the minimum value of this quantity over all $x \geq 0$. [$x=(y^{2} /2) \geq 0$ on the curve]. Equivalently, you can mimimize teh square of teh distance. Differentiate and set the derivative equal to 0. You will find that the minimum is attained at $x=/2$. The minumum value is $\sqrt 2$.

Continuing your method: Instead of considering $y^2 = 2x$, $(1.5,0)$, consider $y=\frac{x^2}{2}, (0,1.5)$ for convenience. The normal line at $(x_0,y_0)$ is: $$y'=x_0 \Rightarrow y=-\frac{1}{x_0}x+y_0+1\Rightarrow y=-\frac{1}{x_0}x+\frac{x_0^2}{2}+1$$ The normal line must pass through $(0,1.5)$: $$1.5=\frac{x_0^2}{2}+1 \Rightarrow x_0=1 \Rightarrow y_0=0.5.$$ The distance between $(0,1.5)$ and $(1,0.5)$ is: $$d=\sqrt{(0-1)^2+(1.5-0.5)^2}=\sqrt{2}.$$

$y^2=2x;$

Point on parabola: $(x,y)$.

Distance^2 to $(1.5,0)$:

$d^2:=(x-1.5)^2+(y-0)^2=$

$(x-1.5)^2+2x= x^2-x+ (1.5)^2=$

$(x-1/2)^2-1/4 +2.25.=$

$(x-1/2)^2 +2;$

Since the square is $\ge 0:$

$d_{min}=√2.$

• Nice alternative Peter! – user Jun 11 '18 at 10:27
• Thanks.Practice in completing the square:)) – Peter Szilas Jun 11 '18 at 11:18

In your implicit differentiation, you got $$\dfrac{dy}{dx} = \dfrac{1}{y_1},$$ so $\dfrac{1}{y_1}$ is the slope of the tangent. The slope of the normal is its negative reciprocal: $$-\dfrac{1}{\frac{1}{y_1}} = -y_1$$ With that, you instead get $y_1 = -y_1(x_1 - 1.5)$ as your point-slope equation. Solve this: \begin{align}y_1 &= -y_1(x_1 - 1.5) \\ 0 &= -y_1(x_1 - 1.5) - y_1 \\ 0 &= y_1(-(x_1 - 1.5) - 1) \\ 0 &= y_1(0.5 - x_1)\end{align} So $y_1 = 0$ or $x_1 = 0.5$. Solve $y_1^2 = 2x_1$ to get the other coordinate (since $(x_1, y_1)$ is on the parabola).

• If $y_1 = 0$, then $x_1 = 0$. The distance between $(0, 0)$ and $(1.5, 0)$ is $1.5$.
• If $x_1 = 0.5$, then $y_1 = \pm 1$. The distance between $(0.5, 1)$ and $(1.5, 0)$ is $\sqrt{2}$. The distance between $(0.5, -1)$ and $(1.5, 0)$ is also $\sqrt{2}$.

The shortest distance among these three possibilities is $\sqrt{2}$.

A reply to an old comment brought me back here, and I noticed that no one had offered a Lagrange multiplier solution. You're looking for the minimum of $$f(x,y) = (x-1.5)^2 +y^2$$ subject to the constraint that $$g(x,y) = 0$$, where $$g(x,y) = y^2-2x$$ The critical points satisfy $$\nabla f(x,y) = \lambda \nabla g(x,y)$$ for some $$\lambda$$. Or, \begin{align*} 2x-3 &= -2\lambda \\ 2y &= 2 \lambda y \end{align*} The second of these implies either $$y=0$$ or $$\lambda = 1$$. If $$y=0$$, then $$x=0$$ and $$f(0,0) = \frac{9}{4}$$. If $$\lambda = 1$$, we can substitute it into the first equation and get $$x = \frac{1}{2}$$. If $$x =\frac{1}{2}$$, then $$y = \pm 1$$, and $$f(\frac{1}{2},\pm1) = 2$$. So the minimum value of $$f$$ is at these two points, and the shortest distance is $$\sqrt{2}$$.