# Finding parametric equations of rectangular equation

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!

• 3 questions: 1) What do you mean by "normal" 2) Are the sides of your rectangle parallel to the coordiante axes ? 3) Could you give us why you are interested in this question, it may help. – Jean Marie Apr 23 '16 at 16:42
• Hey 1. I mean an equation of the form (y = x^2) 2. I dont mean an equation of a rectangle, but just a regular y in terms of x equation with no rotation 3. I am learning about conics and there are a few exercises that ask us to convert the rectangular equation to 2 parametric ones and I can't figure out how – Martin Spasov Apr 23 '16 at 16:46
• Ohhh, I see. It is not the usage to say "rectangular equation", but "cartesian equation" $y=f(x)$. Let us take your example $y=x^2$, the most standard way to convert it into a parametric equation is by setting $x=t, y=t^2$, but $x=t^3, y=t^6$ is another valid solution , etc. as long as relationhip $y$ is the square of $x$ is preserved. – Jean Marie Apr 23 '16 at 16:54
• They were using "rectangular" in the book so i thought that it's ok to use it. Anyway what if I want to find the parametric equation of a hyperbola ? Do i still express y in terms of x and then substitute x with t ? – Martin Spasov Apr 23 '16 at 17:00
• How is this equation expressed ($y=a/x$ ?) – Jean Marie Apr 23 '16 at 17:35

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?