Finding Polar Equation for hyperbola with Graph xy=16

Magnawolf
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Homework Statement



Find a polar equation with the graph as xy=16


Homework Equations



r = ed/(1+- cos\theta)? I'm not really sure at all.

The Attempt at a Solution



I also tried using x = rcos\theta and y = r sin\theta but I can't get anything. I know it's a hyperbola.
 
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Magnawolf said:

Homework Statement



Find a polar equation with the graph as xy=16



The Attempt at a Solution



I also tried using x = rcos\theta and y = r sin\theta but I can't get anything. I know it's a hyperbola.

It is all right. How are r and θ related then?

ehild
 
Inversely? I'm not sure!
 
Magnawolf said:
Inversely? I'm not sure!

No, why do you think so?

Just plug in x=rcosθ and y=rsinθ into the original equation xy=16.

ehild
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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