Nice parameterization of $x^2 + y^2 - kx^2y^2 =1$ Can anyone find a nice simple parameterization of this curve. Just the quarter where $x \ge0$ and $y \ge0$ would be fine.
The parameterization should be "nice" in the sense that the first derivative vector should never be zero or infinite.
You can assume that $0<k<1$.
If it helps, the curve looks somewhat like an circle, but increasing the $k$ parameter makes the curve more square-ish, like a so-called super-ellipse.
Parameterization using rational functions would be very nice, but trigonometric functions would be OK, too.
I've tried the obvious trick: take a line at angle $\theta$ and find its intersection with the curve. This gives a parameterization in terms of $\theta$, but it's a mess. I'm hoping for something simpler.
If it matters to you, this is not homework. This equation represents a part of an aircraft fuselage, and having a parametric representation for it would make certain applications easier. Like drawing it, for example.
 A: If I understand your question right, you just want the part with $0 \leq x,y \leq 1$. Then you could just solve a quadratic equation in either $x$ or $y$ and get something like
$$t \mapsto \left(t,\sqrt{\frac{t^2-1}{kt^2-1}}\right),\quad t \in [0,1].$$
A: Here is the best I could do, after a lot of algebraic fiddling around. Set
$$
w(t) = \frac{2}{1 + \sqrt{1 - k \sin^2 2t}}
$$
and then
$$
x(t) = \sqrt{w}\cos t \quad ; \quad y(t) = \sqrt{w}\sin t
$$
Not too bad, I guess. Maybe the trig functions could be replaced by rational functions, and the whole mess could be simplified ??
A: I'm afraid you'll bump into elliptic curves instead of rational functions. Nonetheless this shows that, 
$$x^2 + y^2 - k x^2 y^2 - 1 = 0\tag{1}$$
has an infinite number of rational solutions. Solving for $y$,
$$y =\frac{\sqrt{1-x^2}}{\sqrt{1-kx^2}}\tag{2}$$
If for some constant $k$ there is rational $x,y$, then one can treat $(2)$ as an elliptic curve. For example, let $k=1/3$, then $(2)$ has solutions $x = 2,\;266/23,\dots$ ad infinitum.
