# Find the cartesian equation from the given parametric equations

I'm tasked with converting these parametric equations into one cartesian equation. $$x = a*sin(t)$$ $$y = b*cos(t)$$

So I begin with my reasoning, which is potentially 100% wrong.

I want to basically solve for t and then substitute. assuming that is a valid thing to do.

$$\frac{x}{a}=sin(t)$$ $$arcsin(\frac{x}{a})=t$$ $$y = b*cos(arcsin((\frac{x}{a}))$$

but now what? is this even acceptable? I'm new to parametric equations and despite searching extensively for answers to this and similar situations I'm having a terrible time making sense of this type of math. Help me math.stackexchange, you're my only hope...

## 2 Answers

Here's a simpler and more elegant way of forming the cartesian equation. \begin{align} bx&=ba\sin(t)\\ ay&=ab\cos(t) \end{align} Squaring we get \begin{align} (bx)^2&=(ba\sin(t))^2\\ (ay)^2&=(ab\cos(t))^2 \end{align} Add the two equations we get \begin{align} (bx)^2+(ay)^2&=(ba\sin(t))^2+(ab\cos(t))^2\\ &=(ab)^2(\sin^2(t)+\cos^2(t))\\ &=(ab)^2 \end{align} Simplifying by dividing both side by $(ab)^2$. $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Hint: Assuming $a\neq 0 , b\neq 0\implies \dfrac x a = \sin t \text { and } \dfrac y b = \cos t$.

There is a fundamental equation involving just $\sin t$ and $\cos t$. Apply it and you get the equation in $x,y$ - coordinates.

• $$cos^2(t)+sin^2(t)=1$$ – James Sep 28 '14 at 19:00
• Exactly! That's the fundamental equation! – thanasissdr Sep 28 '14 at 19:01