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Is this parameterisation an ellipse: \begin{align}x(t) &= \frac{2 \cos(t)}{1 + a \sin(t)}\\ y(t) &= \frac{2 \sin(t)}{1 + a \sin(t)}\end{align} where $a$ is a real positive parameter.

I tried to do it the naive way but couldn't find a definitive answer.

Plotting our curve with the help of Geogebra gives the following very ellipse like graph:

Plotting our curve with the help of Geogebra gives the following very ellipse like graph.

Any help would be appreciated.

Martin Sleziak
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Richard Kruel
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  • Nice picture. I think it's unlikely to be an ellipse. You probably want $a < 1$ to avoid a $0$ in the denominator. – Ethan Bolker Jun 12 '17 at 14:43
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    Using your parameterisation, can you form $x^2(t)+y^2(t)$? If so, does that look like the cartesian equation of an ellipse? –  Jun 12 '17 at 14:44
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    Welcome to StackExchange! When you say you tried the naive way, what exactly do you mean? Showing your working my help people point out if you have gone wrong or help you understand things you are confused about – lioness99a Jun 12 '17 at 14:44
  • We formed x(t)^2+y(t)^2 but didn't think of eliminating t.. However the problem is now settled. Thank you guys! – Richard Kruel Jun 12 '17 at 15:09
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    An explanation of the diagram would be nice. What are the green, red, and black curves? What is $a$? – TonyK Jun 13 '17 at 13:45
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    @RichardKruel Be sure to accept an answer below if it answered your question. – Bonnaduck Jun 13 '17 at 17:22

5 Answers5

45

It's easy to see that

$$x^2+y^2={4\cos^2t\over(1+a\sin t)^2}+{4\sin^2t\over(1+a\sin t)^2}={4\over(1+a\sin t)^2}$$

(as Bacon found). But also

$$y={2\sin t\over1+a\sin t}\implies ay={2a\sin t\over1+a\sin t}={2(1+a\sin t)-2\over1+a\sin t}=2-{2\over1+a\sin t}$$

and thus

$${2\over1+a\sin t}=2-ay$$

hence

$$x^2+y^2=(2-ay)^2=4-4ay+a^2y^2$$

or

$$x^2+(1-a^2)y^2+4ay=4$$

As lhf points out, this is an equation for an ellipse, parabola, or hyperbola depending on the sign of $a^2-1$.

Barry Cipra
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25

Asking WA to eliminate $t$ gives $a^2 y^2 - 4 a y - y^2 + 4 = x^2$. Therefore, we have

  • an ellipse if $a^2-1<0$

  • a parabola if $a^2-1=0$

  • a hyperbola $a^2-1>0$

lhf
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$$\begin{align} x&=\frac {2\cos t}{1+a\sin t}\tag{1}\\ y&=\frac {2\sin t}{1+a\sin t}\tag{2}\\ (2)/(1):\hspace{3cm}\\ \frac yx&=\tan t\tag{3}\\ (1)^2+(2)^2:\hspace{3cm}\\ x^2+y^2&=\frac 4{(1+a\sin t)^2}\\ &=\frac {4(x^2+y^2)}{\big(\sqrt{x^2+y^2}+ay\big)^2} &&\scriptsize \bigg(\sin t=\frac y{\sqrt{x^2+y^2}}\bigg)\\ (x^2+y^2)\big[\big(\sqrt{x^2+y^2}+ay\big)^2-4\big]&=0\\ \because{x^2+y^2}\neq 0\therefore \qquad \big(\sqrt{x^2+y^2}+ay\big)^2-4&=0\\ x^2+y^2&=(\pm2-ay)^2\\ \color{red}{x^2+(1-a^2)y^2\pm 4ay-4}&\color{red}{=0} \end{align}$$ which is an ellipse if $a^2<1$, per criteria outlined here.

Hypergeometricx
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$$x^2 + y^2 = \frac{4}{(1+a \sin t)^2}$$

Now substitute in for $\sin t$ and place bounds on $a$ to correctly find the cartesian equation of an ellipse.

3

Renaming $t$ as $\theta$, you have a polar equation:

$$\rho=\frac2{1+a\sin\theta}.$$

This is the polar form of the equation of a conic with a focus at the origin. https://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_focus

If $|a|<1$, the denominator has no root hence the curve is bounded.

By rewriting

$$\rho+ay=2$$ you establish

$$x^2+y^2=(2-ay)^2.$$

This conic is an ellipse when the discriminant of the quadratic terms $x^2+(1-a^2)y^2$ is negative, i.e. when $|a|<1$.

  • [+1] I like your answer. It's by far the simplest of all. A little remark : the last formula you give, written under the form $\sqrt{x^2+y^2}=|a| |\tfrac{1}{a}-y|$ expresses the focus-directrix definition of an ellipse. – Jean Marie Jan 22 '18 at 20:16