# Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$.

Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.

What I did so far is :

plot the y var from $x^2-y^2=1 \implies y=\sqrt{1-x^2}$

Create a distance equation : $d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$d = \sqrt{x^2+(1-y)^2}$

$d = \sqrt{2-2\sqrt{1-x^2}}$

$\displaystyle \frac d{dx} (d) =\frac x{\sqrt{1-x^2}\sqrt{2-2\sqrt{1-x^2}}}$

I need to find the max or min? Any suggestions? thanks!

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$\LaTeX$ hints: To get the square root sign, put a slash before the sqrt and braces around the stuff that goes inside: \sqrt{1+x^2} gives $\sqrt{1+x^2}$ For stacked fractions, use \frac {numerator}{denominator} so \frac{1+x^2}3 gives $\frac{1+x^2}3$ You can also get line breaks inside $\LaTeX$ with a double backslash. +1 for showing your work –  Ross Millikan Apr 17 '13 at 13:32

Your approach is fine. It is a little easier to minimize $d^2$ than $d$ as that gets rid of one square root. You lost a sign when you went from $d=\sqrt{x^2+(1-y)^2}$ to $d=\sqrt{2-2\sqrt{1-x^2}}$ as the $y^2$ term is positive. Plotting the curve shows that the distance to either branch from $(0,1)$ is the same, so we can use the positive square root $x=\sqrt{1+y^2}$. We have $$d^2=x^2+(1-y)^2=2-2y+2y^2\\\frac {d(d^2)}{dy}=-2+4y$$ which we set to zero and find $y=\frac 12, x=\sqrt{\frac 54}$. The squared distance is then $d^2=\frac 54+\frac 14=\frac 32$ and the linear distance is $d=\sqrt \frac 32$. This must be a minimum, as there are points on the hyperbola very far away.

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if the distance is sqrt of 3/2 how i know the point that have this distance from P? –  Ofir Attia Apr 17 '13 at 13:41
@OfirAttia: I got that first. It is $(\sqrt \frac 54,\frac 12)$. There is also $(-\sqrt \frac 54,\frac 12)$ at the same distance. –  Ross Millikan Apr 17 '13 at 14:04

Assume $f\colon R^2\mapsto R$ is $C^1$, $P(x_0,y_0)$, $c\in R$ and $f^{-1}(c)\neq\emptyset$. For obvious reasons there is a point $Q\in f^{-1}(c)$ of minimal distance from $P$. Clearly $PQ$ is normal to $f^{-1}(c)$. So we have to solve $$\frac{y-y_0}{x-x_0}=\frac{f_y}{f_x}.$$

In our case: $f(x,y)=x^2-y^2$, $c=1$, $x_0=0$, $y_0=0$. Thus $$\frac{y-y_0}{x-x_0}=\frac{f_y}{f_x}\iff\frac{y-1}{x}=-\frac{y}{x}\iff y=\frac{1}{2},$$ it follows that $x=\pm\sqrt{\frac{5}{4}}$, from which the distance is easily computed.

Moral: It seems to be a deviation not to compute the distance directly instead to compute the point where it shows up, but it actually isn't; the lengthy computations in the other answers shows that clearly.

Michael

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Any point$(Q)$ on the hyperbola $x^2-y^2=1$ can be represented as $(\sec t,\tan t)$

The distance of $Q(\sec t,\tan t)$ from $P(0,1)$ is $$\sqrt{(0-\sec t)^2+(1-\tan t)^2}=\sqrt{2-2\tan t+2\tan^2t}=\sqrt{\frac{(2\tan t-1)^2+3}2}\ge \sqrt{\frac32}$$

We can also apply Second derivative test on $\sqrt{2-2\tan t+2\tan^2t}$ to find the minimum distance

$\sqrt{2-2\tan t+2\tan^2t}$ will be minimum iff $2-2\tan t+2\tan^2t$ is minimum positive

Let $f(t)=2-2\tan t+2\tan^2t$

$f'(t)=-2\sec^2t+4\tan t\sec^2t=2\sec^2t(2\tan t-1)$

For the extreme values of $f(t),f'(t)=0$

$\implies \sec^2t(2\tan t-1)=0$ $\implies 2\tan t-1=0$ as $\sec^2t\ge1$

Now, $f''(t)=2\sec t(\sec t\tan t)(2\tan t-1)+\sec^2t(2\sec^2t)$

$$\text{ At }\tan t=\frac12, f''(t)=2\sec^4t>0$$

$$\text{ At }\tan t=\frac12,f(t)=\frac32$$

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why i cant represent point with(cos,sin)? –  Ofir Attia Apr 17 '13 at 9:00
@OfirAttia, as $\sec^2-\tan^2t=1$ but $\cos^2t+\sin^2t=1$ –  lab bhattacharjee Apr 17 '13 at 9:04
i think the second derivative is 0 here –  Ofir Attia Apr 17 '13 at 11:52
@OfirAttia, please find the edited answer –  lab bhattacharjee Apr 17 '13 at 12:55
Sorry for the ignorance but I do not understand how I use sec. in my class we dont use it at this moment so i dont know how to use it and i dont know if my professor will agree to this answer –  Ofir Attia Apr 17 '13 at 13:08