The shortest distance from a point to the graph of the function

Question

To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$. I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?

Answer 1

You're after the minimum of $f(x)=(x-5)^2+\left(\frac4x-5\right)^2$, with $x\in\mathbb R$. Now,\begin{align}f'(x)&=-10-\frac{32}{x^3}+\frac{40}{x^2}+2x\\&=2\frac{x^4-5x^3+20x-16}{x^3}\\&=2\frac{(x^2-4)(x^2-5x+4)}{x^3}.\end{align}So, see what happens at the roots of $f'(x)$, which are $\pm2$, $1$ and $4$.

Answer 2

I offer this as an alternative solution.

Looking at the graph of $xy=4$ and the point $P=(5,5)$, clearly we need only consider the branch of $xy=4$ that is in the first quadrant. So we need only consider the function $y = \dfrac 4x$ where $x > 0$.

Consider the particular point $Q=\left(\xi, \dfrac{4}{\xi}\right)$ on the curve $y = \dfrac 4x$. The slope of the tangent to the curve at the point $Q$ is $y'=-\dfrac{4}{\xi^2}$. So the slope of the normal line to the curve at that point is $\dfrac 14 \xi^2$. The equation of the normal line to the curve at point $P$ is therefore

$$y-\dfrac{4}{\xi} = \dfrac 14\xi^2(x- \xi)$$

$Q$ will be the closests point on the curve to the point $P=(5,5)$ when the normal line contains the point $Q$. So

\begin{align} 5-\dfrac{4}{\xi} &= \dfrac 14\xi^2(5- \xi) \\ 20\xi -16 &= \xi^3(5-\xi) \\ 20\xi -16 &= 5\xi^3-\xi^4 \\ \xi^4 - 5\xi^3+20\xi-16 &= 0 \\ \xi^4 - 5\xi^3 + 4\xi^2 - 4\xi^2 + 20\xi - 16 &= 0 \\ \xi^2(\xi^2 - 5\xi + 4) - 4(\xi^2 - 5\xi + 4) &= 0 \\ (\xi^2 - 4)(\xi^2 - 5\xi + 4) &= 0 \\ (\xi + 2)(\xi - 2)(\xi - 1)(\xi - 4) &= 0 \\ \xi &\in \{ 1,2,4 \} \end{align}

\begin{array}{c} \xi & Q & d(P,Q) \\ \hline 1 & (1,4) &\sqrt{17} \\ 2 & (2,2) & \sqrt{18} \\ 4 & (4,1) & \sqrt{17} \\ \hline \end{array}

So the points on the curve $xy=4$ that are closest to the point $(5,5)$ are $(1,4)$ and $(4,1)$ and the distance is $\sqrt{17}$.