1
$\begingroup$

enter image description here

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

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

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 $(a\cosh(t),b\sinh(t))$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ if it is not in the center? $\endgroup$ – Harry Oct 26 '13 at 3:51
  • $\begingroup$ @Harry revised to answer that. $\endgroup$ – J. W. Perry Oct 26 '13 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.