Finding parametric equations of rectangular equation

Question

Is there a general process to follow when finding the parametric equations of a normal rectangular equation ? I know that one rectangular equation might have many parametric equations, but are there some steps that might help me ? I looked at Finding Parametric Equations For A Rectangular Equation, but it did not help a lot. I am learning PreCalc so please if you use methods taught in Calc make sure to explain them well. Thank you!

Answer 1

I think that this question requires some clarification.

First of all we can properly talk about the parametric equation of a curve, not of a function.

A function, as $f: X \to \mathbb{R} \quad y=f(x) \quad X \subset \mathbb{R} \quad$ has a graph in $\mathbb{R}^2$ that is a curve and this curve can be represented, parametrically, as the graph of a function $p(t):I \to \mathbb{R}^2$ where $I$ is an interval in $\mathbb{R}$.

This can be done using every bijective function $g:I \to X$ so that we have a function: $$ p:I \to X\times \mathbb{R} \qquad p(t)=(g(t),f(g(t))) $$

As an example, if we have $f: \mathbb{R} \to \mathbb{R} \quad y=ax$, we can use $g:(-\pi/2,\pi/2)\to \mathbb{R}\quad g(t)=\tan t$ and we have: $$ p(t):(-\pi,\pi) \to \mathbb{R}^2 \quad p(t)=(\tan t, a \tan t) $$ note that we can use any bijective function from an interval to $X$

But this is not the only way to obtain a parametric representation of a curve. To show some example:

For the function $f:[-1,1]\to \mathbb{R} \quad y=f(x)=\sqrt{1-x^2}$, we can use a parametric representation :

$$ p:[0,\pi] \to \mathbb{R} \quad p(t)=(\cos t, \sin t) $$ That can be found using the previous method and the trig. identity $\sin^2 x+\cos^2 x=1$.

For an hyperbola of equation $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$ we have not an equation of the form $y=f(x)$ but, since it is a curve in $\mathbb{R}^2$, we can have the parametric equation: $$ p(t)=(a \sec t, b \tan t) \quad t \in [-\pi, \pi] $$ that is derived from the trigonometric identity $ \sec^2 \alpha - \tan^2 \alpha=1$.

Finally note that another parametric representation can also be write in the form: $$ x=a\frac{1+t^2}{1-t^2} \quad y=b\frac{2t}{1-t^2} $$

can you see from where this come from?