Is this an ellipse? 
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: 
 
Any help would be appreciated.
 A: $$\begin{align}
x&=\frac {2\cos t}{1+a\sin t}\tag{1}\\
y&=\frac {2\sin t}{1+a\sin t}\tag{2}\\
(2)/(1):\hspace{3cm}\\
\frac yx&=\tan t\tag{3}\\
(1)^2+(2)^2:\hspace{3cm}\\
x^2+y^2&=\frac 4{(1+a\sin t)^2}\\
&=\frac {4(x^2+y^2)}{\big(\sqrt{x^2+y^2}+ay\big)^2}
&&\scriptsize \bigg(\sin t=\frac y{\sqrt{x^2+y^2}}\bigg)\\
(x^2+y^2)\big[\big(\sqrt{x^2+y^2}+ay\big)^2-4\big]&=0\\
\because{x^2+y^2}\neq 0\therefore \qquad
\big(\sqrt{x^2+y^2}+ay\big)^2-4&=0\\
x^2+y^2&=(\pm2-ay)^2\\
\color{red}{x^2+(1-a^2)y^2\pm 4ay-4}&\color{red}{=0}
\end{align}$$
which is an ellipse if $a^2<1$, per criteria outlined here.
A: $$x^2 + y^2 = \frac{4}{(1+a \sin t)^2}$$
Now substitute in for $\sin t$ and place bounds on $a$ to correctly find the cartesian equation of an ellipse.
A: 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 $a^2-1$.
A: 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 $|a|<1$.
A: 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 $a^2-1>0$
