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


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

  • 3
    $\begingroup$ Nitpick: how do you define the "distance" between an equation and a point? (You really mean the distance between the parabola and the point.) $\endgroup$ – Andreas Rejbrand Jun 11 '18 at 11:05

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


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

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

  • 2
    $\begingroup$ 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. $\endgroup$ – Paŭlo Ebermann Jun 11 '18 at 21:09
  • $\begingroup$ @PaŭloEbermann Yes of course by symmetry, indeed the distance for y positive or negative is the same $$d=\sqrt{(y-0)^2+\left(x-\frac32\right)^2}=\sqrt{(-y-0)^2+\left(x-\frac32\right)^2}$$ $\endgroup$ – gimusi 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$$\left\|\left(\frac12,1\right)-\left(\frac32,0\right)\right\|=\sqrt2.$$ enter image description here

  • $\begingroup$ +1 for picture. $\endgroup$ – Matthew Leingang Jun 11 '18 at 12:28

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: $$d=\sqrt{(0-1)^2+(1.5-0.5)^2}=\sqrt{2}.$$


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 $\sqrt 2$.



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

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


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

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

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

Since the square is $\ge 0:$


  • 1
    $\begingroup$ Nice alternative Peter! $\endgroup$ – gimusi Jun 11 '18 at 10:27
  • $\begingroup$ Thanks.Practice in completing the square:)) $\endgroup$ – 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 $\sqrt{2}$.


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