Eliminate the parameter to find a description of the following circles or circular arcs in terms of $x$ and $y$. Give the center and radius, and indicate the positive orientation.

$x=4\cos{(t)} ,\ y=3\sin{(t)} ;\ 0 \leq t \leq 2\pi$


$\displaystyle x^2=4^2\cos^2{(t)} ,\ y^2=3^2\sin^2{(t)} \implies \frac{x^2}{4^2}=\cos^2{(t)} ,\ \frac{y^2}{3^2}=\sin^2{(t)}$

But I detect no radius. I'm rather confused, on this whole question. It doesn't even explicitly define "the parameter". Insight?

  • 1
    $\begingroup$ I might not understand the question, but $(4\cos(t),3\sin(t))$ does not describe a circle (it is an ellipse instead). $\endgroup$ – Hirshy Jul 28 '15 at 7:15

Those equations describe an ellipse, rather than a circle. I'm guessing that by radius they mean length of major and minor axes?

The parameter is $t$, and you're on the right track to eliminate it; your next step is to add the equations

$$ \frac{x^2}{4^2}=\cos^2{(t)} \\ \frac{y^2}{3^2}=\sin^2{(t)}$$

and use the pythagorean identity.

  • $\begingroup$ Such that $\displaystyle \frac{x^2}{4^2}+\frac{y^2}{3^2}=\cos^2{t}+\sin^2{t} \implies \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$? Yeah? $\endgroup$ – alxmke Jul 28 '15 at 7:21
  • $\begingroup$ Yup, that's it, I think. $\endgroup$ – coldnumber Jul 28 '15 at 7:22
  • $\begingroup$ What do you think could be implied by the prompt "indicate the positive orientation"? $\endgroup$ – alxmke Jul 28 '15 at 7:32
  • $\begingroup$ Are you also asked to graph it? See here: mathworld.wolfram.com/CurveOrientation.html (you travel from $t=0$ to $t=2\pi$) $\endgroup$ – coldnumber Jul 28 '15 at 7:34
  • $\begingroup$ I am not, but it asks for the elimination of $t$, so would it still be referencing $t$ even after asking for its elimination? If the parameters were given as a parametric equation, then for $t$ going $0\to2\pi$, the positive orientation would be ... counterclockwise, no? $\endgroup$ – alxmke Jul 28 '15 at 7:38

You have not a radius since you are describing an ellipse. Indeed

$$ 1=cos^2(t)+sin^2(t)=\frac{x^2}{16}+\frac{y^2}{9}, $$ then


The major semi axis is $4$, and the minor semi axis in $3$.


Notice, $$x=4\cos t \implies \cos t=\frac{x}{4}\tag 1$$ & $$y=3\sin t \implies \sin t=\frac{y}{3}\tag 2$$ Now, for eliminating $t$, squaring & adding (1) & (2), we get $$\cos^2t+\sin^2t=\left(\frac{x}{4}\right)^2+\left(\frac{y}{3}\right)^2$$ $$\color{blue}{\frac{x^2}{16}+\frac{y^2}{9}=1}$$ The above equation is in the form of the standard form of equation of an ellipse: $\color{blue}{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ where, $a=4$ & $b=3$.

Hence, the curve represents an ellipse not a circle, hence we have $$\text{center of ellipse, origin}\equiv(0, 0)$$ $$\text{major axis}, 2a=2\times 4=8$$ $$\text{minor axis}, 2b=2\times 3=6$$ O.P.'s detection is right that there is no radius as that is not a circle but the curve represents an ellipse.


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