If we have an ellipse with equation $x^2/a^2+y^2/b^2=1$ then if we were to change this into polar coordinates then would the parametrisation be $x=ar\cos(\theta)$ and $y=br\sin(\theta)$?
Also what would the parametrisation of a hyperbola be?
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2The point $(a\cos\theta,b\sin\theta)$ is not at angle $\theta$ from the $x$-axis. This appears to be a common mistake. – May 06 '14 at 16:28
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Parameterization and Conversion to Polar coordinate are not linked
To parameterize hyperbola $$\frac{(x-\alpha)^2}{a^2}-\frac{(y-\beta)^2}{b^2}=1\text{ and }\sec^2\phi-\tan^2\phi=1$$
$$\frac{x-\alpha}a=\sec\phi, \frac{y-\beta}b=\tan\phi$$
For Conversion to Polar coordinate always $$x=r\cos\theta,y=r\sin\theta$$
lab bhattacharjee
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ok so if we parametrise a surface integral to evaluate it we don't multiply by the Jacobian as we would to if we had changed it to polar coordinates? – user134785 May 06 '14 at 17:11
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For a hyperbola, you can also use hyperbolic functions, since $\mathrm{ch}^2 t - \mathrm{sh}^2 t = 1$. – Jean-Claude Arbaut May 06 '14 at 17:37