why do we always take the parametre of parametric equation of ellipse as the angle formed with x-axis instead of semi-major axis?
Solution 1:
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:
- An ellipse is vertically oriented if $b > a$, and
- 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.