# Shortest distance from point to curve

I could use some help solving the following problem. I have many more like this but I figured if I learn how to do one then I can figure out the rest on my own. Thanks in advance!

A curve described by the equation $y=\sqrt{16x^2+5x+16}$ on a Cartesian plane. What is the shortest distance between coordinate $(2,0)$ and this line?

• Can you elaborate on what it means for a line to follow a curve? Is the question simply to find the shortest distance between $(2,0)$ and that curve? Commented May 3, 2017 at 22:28
• the equation of the line. ill edit to clear that up. thx Commented May 3, 2017 at 22:29
• yes, the solution would be to find the shortest distance between the point (2,0) and the curve Commented May 3, 2017 at 22:32

Start by finding the distance from some point on the curve to $(2,0)$ in terms of $x$. Using the distance formula, we get $$D=\sqrt{(x-2)^2+(\sqrt{16x^2+5x+16}-0)^2}$$ $$D=\sqrt{x^2-4x+4+16x^2+5x+16}$$ $$D=\sqrt{17x^2+x+20}$$ This will end up being a messy derivative. However, since the distance $D$ will never be negative, we can minimize $D^2$ instead of $D$ and still get the same answer. So now we get $$D^2=17x^2+x+20$$ $$\frac{dD^2}{dx}=34x+1$$ Now we set this equal to $0$ and solve for $x$: $$34x+1=0$$ $$x=-\frac{1}{34}$$ So the distance is minimized at $x=-\frac{1}{34}$, and to find the minimum distance, simply evaluate $D$ when $x=-\frac{1}{34}$.

• Let's suppose that D can be negative, then can we still simplify using $D^2$? Commented Apr 11, 2018 at 6:47
• @PianoLand No, but there is something wrong with your D formula if you can't do that. Commented May 27, 2021 at 2:46

Since distance is positive and the square root function is increasing, it suffices to find the smallest value the squared distance between $(x,y)$ on the curve and the point $(2,0)$ can take. This is $$L(x) = (x-2)^2 + (y-0)^2 = (x-2)^2+y^2 = x^2-4x+4 + 16x^2+5x+16 = 17x^2+x+20.$$ A minimum can only occur if $L'(x)=0$. So $$L'(x) = 34x+1,$$ so there is a turning point at $x=-1/34$. Moreover, the derivative is negative on the left and positive on the right, so the point is a minimum. Hence the minimum distance is $$\sqrt{L(-1/34)} = \sqrt{\frac{1359}{68}} \approx 4.47.$$

• Why not a negative $x$? Commented May 3, 2017 at 22:50
• Brain fart. Thanks for pointing that out... Commented May 3, 2017 at 22:56
• Truly an anomaly from someone with a 33.5k reputation! Commented May 3, 2017 at 22:57
• Let us hope so! Commented May 3, 2017 at 23:06

Hint 1: take a point $(x, y)$ on the curve, calculate its distance from $(2, 0)$. The fact that the point is on the curve allows you to express that distance in terms of $x$ alone. Then find the minimum (but check the second hint first).

Hint 2: instead of minimizing the distance, minimize the square of the distance.