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...


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.

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

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