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For ellipses having equation in the form of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ why are the parametric equations are always $$x=a\cos(\theta)$$ $$y=b\sin(\theta)$$ even when $b>a$?

As far as I know, for hyperbola having equation
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm 1$$the parametric equations are $(x(\theta),(y(\theta))=(a\sec(\theta),b\tan(\theta))$ and $(x(\theta),(y(\theta))=(a\tan(\theta),b\sec(\theta))$ for + and - respectively. Which I suppose that the angle is considered with the transverse axis. But why is it always the angle with $x$-axis for parametric equations of ellipse?

Thank you.

Red Five
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MSKB
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  • The $(x,y)=(a\cos\theta,b\sin\theta)$ parameterization of the ellipse "doesn't care" that the curve is an ellipse; there's no consideration for foci and eccentricity and whatnot. It considers the curve merely a stretched circle, with arbitrary vertical and horizontal stretching, in order to leverage the $\sin^2\theta+\cos^2\theta=1$ identity. The fact that a stretched circle is an ellipse is extra knowledge. – Blue Dec 17 '21 at 14:47
  • You can contrast with the polar parameterization: $$r = \frac{p}{1-e\cos\theta}$$ (You can Cartesian-ify the parameterization by taking $(x,y)=(r\cos\theta, r\sin\theta)$.) This parameterization works for any conic, and "cares" about the conic-sectional nature of the curve: it assumes the major/transverse axis aligns with the $x$-axis (and that a focus coincides with the origin); $p$ is the length of the latus rectum. Replacing $\theta$ with $\theta-\theta_0$ rotates the curve; still, how points are traced is still effectively governed by an angle made with the major/transverse axis. – Blue Dec 17 '21 at 15:08
  • Parameter $\theta$ in those equations is NOT the angle formed by ray $OP$ with $x$ axis. And if you want you can write the equations exchanging $\cos$ and $\sin$, it doesn't matter. – Intelligenti pauca Dec 17 '21 at 15:37

1 Answers1

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The parametrization of the ellipse with an equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ will be $$\begin{cases}x = a\cos\theta \\ y = b\sin\theta\end{cases}$$ for $\theta \in [0, 2\pi]$. If you need an explanation for this, see my answer in a similar question.

Given this equation, we now define that:

  1. An ellipse is vertically oriented if $b > a$, and
  2. An ellipse is horizontally oriented if $b < a$.

I think you are confusing the orientation of an ellipse with its equation. We first get the equation, and just define a particular property satisfying a condition. As Blue commented,

The $(x,y) = (a \cos\theta, b\sin\theta)$ parameterization of the ellipse "doesn't care" that the curve is an ellipse; there's no consideration for foci and eccentricity and whatnot. It considers the curve merely a stretched circle, with arbitrary vertical and horizontal stretching, in order to leverage the $\sin^2\theta + \cos^2\theta=1$ identity. The fact that a stretched circle is an ellipse is extra knowledge.

In the case of a hyperbola with an equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1,$$ the parametrization matters because it is a difference, not a sum. The difference of two terms here is not commutative. Try replacing $(x(\theta),(y(\theta))=(a\sec(\theta),b\tan(\theta))$ in the equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = - 1$$ and see if it still holds.

soupless
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  • I was not confusing the orientation of ellipse I know that for b>a the major axis will be the y-axis. But the thing I didn't understand that why don't we usually take the angle formed by the ray with the y-axis for b>a instead of the angle formed with x-axis. While practicing few problems I noticed that in every websites/pages the parametric equations I have mention led in my question were used whereas I solved them by exchanging sine and cosine keeping other terms the same – MSKB Dec 17 '21 at 15:46
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    @MSKB Sorry for assuming that you were. You can do that too. Taking the angle formed by the ray with the $y$-axis in a counterclockwise manner is the same as if you did the same with the $x$-axis with an added $\pi/2$. – soupless Dec 17 '21 at 15:57