# Finding the perimeter of the shape of $x^2+x^2y^2+y^2=a^2$ [closed]

I need to find the perimeter of the shape of $$x^2+x^2y^2+y^2=a^2$$ ($$a$$ is any real number) in the Cartesian plane.

Is there any general formula that depicts the ratio of area to perimeter of every given shape?

• Please read the following post on meta, so you can start improving the quality of the questions you ask: How to ask a good question (on math.se). Mar 7, 2020 at 23:01
• @aminabzz No general formula, it can be from area and arc length squared by integration Mar 8, 2020 at 11:36
• No comment on my solution ? Mar 8, 2020 at 21:18

Not a full answer, but a help...

First of all, trying to find the ratio of area to perimeter is like multiplying difficulties. In particular, this ratio isn't a constant in general.

Here is how I have considered the issue.

I have begun by a plot of the curves for increasing values of parameter $$a$$ (see figure below, the largest curve being for $$a=4$$).

Fig. 1 : The curves are convex till $$a=2$$, non-convex beyond.

The equation is given in implicit form $$f(x,y)=0$$ ; on this equation, we see different things : $$\underbrace{x^2+y^2}_{\text{dominant for small}\ a}+\underbrace{x^2y^2}_{\text{dominant for large}\ a}=a^2$$

for example, the fact that the curves are like circles for small $$a$$ and like hyperbolas for large values of $$a$$; the fact that they are invariant for different symmetries ($$x \leftrightarrow -x, y \leftrightarrow -y, x \leftrightarrow y$$) that we evidently find back on the curves. Consequence : it suffices to study theses curves in the first quadrant (the first half of the first quadrant would even be enough).

We can switch to an equivalent cartesian equation (exercise!),

$$y:=\pm f_a(x)=\pm \sqrt{\dfrac{a^2-x^2}{1+x^2}}$$

well, a pair of cartesian equations.

From here, you can compute the length of $$1/4$$ of the curve by a classical formula :

$$L_a=\int_0^{a} \sqrt{1+f'_a(x)^2}dx=\int_0^{a}\sqrt{1+\dfrac{x^2(a^2 + 1)^2}{(x^2 + 1)^3(a^2 - x^2)}}dx$$

It seems one cannot obtain an exact expression for $$L_a=L(a)$$. Numerical computations give the following curve, with good approximation by

$$M(a)=a \dfrac{6a-1}{\pi a+1}$$

(thanks to @Blue who has indicated that my first approximation wasn't exact).

Fig. 2 : Evolution of perimeter $$L=L_a$$ (Blue) as a function of $$a$$ and its approximation $$M$$ (Red). Interpretation : for $$a=500$$, one finds $$L(a)=962.2$$ giving a total perimeter $$4 \times 962.2$$

• I agree with your work ... up until the "linear relationship" part. While $L_a$ does seem close to $a\pi/2$ for a while, and the overall plot of $L_a$ as a function of $a$ does seem remarkably "straight", after about $a=13$ or so, the values seem to start converging on $2a$.
– Blue
Mar 8, 2020 at 8:21
• I personally was struck by how the curve's equation could be re-written $$(1+x^2)(1+y^2) = 1+a^2$$ which would seem to invite defining $x=\tan\theta$, $y=\tan\phi$, $a=\tan\alpha$ so that (in the first quadrant) $$\sec\theta\sec\phi=\sec\alpha$$ But it doesn't really make perimeter calculation any less complicated. It would help for OP to explain the "need" for finding this particular perimeter —eg, is it a textbook exercise or just a personal curiosity?— to give a sense of what level of complexity should be anticipated (or tolerated) here.
– Blue
Mar 8, 2020 at 8:32
• @Blue Thanks for having taken the time to have a look at this issue, and for your different remarks. I thought these kind of quartics are known and listed somewhere but I haven't found anything. You are right, the OP should give more information. Mar 8, 2020 at 10:08
• Jean, good. I started this but did not get as far as you. I got just far enough to bet that there was no closed form. Mar 9, 2020 at 16:50

One can start from $$y= f(x)=\sqrt{\frac{a^2-x^2}{1+x^2}}$$

A quarter of the star parametrized on a single $$u, (0 per Blue's comment:

$$\tan \alpha= \sqrt{1+a^2},\, (x,y)= ({\tan \alpha \tan u -1},{\tan\alpha \cot u -1}\,)$$

$$r^2 = 2 ( \frac{\tan \alpha}{\sin 2u}-1)$$
and polar angle $$\theta$$ (please check)
$$\tan^2\theta=\frac{\sin(u-\alpha)}{\cos(u+\alpha)}.$$
• I have not worked it out, the $\theta$ dependence. Mar 8, 2020 at 19:46