Find Closest Point on Hyperbola: xy=8 to (3,0)

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1) Find the point on the hyperbola xy=8 closest to (3,0).

I honestly, have no idea what to do. I seriously do not remember discussing anything like this in class, nor having any previous problems in homework. If anyone can give me a start or walkthrough, that would be fantastic!
 
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Well, y = 8/x gives you a relation between the two variables that allows you to write the distance from (x,y) to (3,0) in terms of one variable, which can then be easily minimized.
 
I've got D = 2rtx^2 - 6x + 9 + 64/x^2

I'm not sure as to how this minimizes?
 
HOW did you get that? What are r and t? What is D? There were not "r", "t", or "D" in your orginal statement of the problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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