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.

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 $a^2-1$.

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 $a^2-1>0$

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