<s>[INST] 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!
 [/INST] 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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