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Any point on an ellipse can be wrttien as $(a\cos\theta,b\sin\theta)$, How could we genarilse this to a hyperbola?

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  • $\begingroup$ $(a\sec(\theta), b\tan(\theta))$? Perhaps you will have to move this around a bit for the most general formula. $\endgroup$ Commented Oct 26, 2013 at 3:47
  • $\begingroup$ @ Harry: mathworld.wolfram.com/Hyperbola.html (see equation 21 and 22) $\endgroup$
    – Amzoti
    Commented Oct 26, 2013 at 3:47
  • $\begingroup$ @Amzoti is that $a,b$ have similar meaning like ellipse $\endgroup$
    – Harry
    Commented Oct 26, 2013 at 3:50
  • $\begingroup$ See equations 13 and 14 mathworld.wolfram.com/Ellipse.html $\endgroup$
    – Amzoti
    Commented Oct 26, 2013 at 3:51

3 Answers 3

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Alternately, since this is a hyperbola after all, use hyperbolic trig functions: $(a\cosh(t),b\sinh(t))$. The hyperbolic analogue of the Pythagorean identity is $\cosh^2 - \sinh^2 = 1$, so the curve satisfies $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The parameter $t$ can be interpreted as hyperbolic angle, the area swept out by a ray from the origin to the curve as it traverses from the vertex of the hyperbola to $(a\cosh(t),b\sinh(t))$.

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I know you asked for a transcendental parametrization, but don’t forget the rational parametrizations of the circle $X^2+Y^2=1$ such as $X=2t/(t^2+1)$, $Y=(t^2-1)/(t^2+1)$.

Similarly, if you want to parametrize the hyperbola $Y^2-X^2=1$, you can use $X=2t/(t^2-1)$, $Y=(t^2+1)/(t^2-1)$. In both cases, there are many other similar parametrizations that you can use.

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You are probably looking for $(a \sec \theta,b\tan \theta)$ if centered at the origin like your ellipse form. Note that I have applied no effort to answer this, here is a pretty decent expnanation.

If it is not centered at the origin, then the parametric form for the hyperbola, $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$ is $$(h+a\sec \theta, k+b\tan \theta).$$

Again, look at that link, and do edit your post if you want some sort of detailed derivation.

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  • $\begingroup$ if it is not in the center? $\endgroup$
    – Harry
    Commented Oct 26, 2013 at 3:51
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    $\begingroup$ @Harry revised to answer that. $\endgroup$ Commented Oct 26, 2013 at 3:58

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