Why do we always take the parameter of parametric equation of ellipse as the angle formed with x-axis instead of semi-major axis

conic sections

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.

Best Answer

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.

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