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 $P$.

Question

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!

Answer 1

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.

Answer 2

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

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

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

Answer 3

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