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Does anyone know how to sketch $y^2=x^2(a^2-x^2)?$ Here $a$ is a constant.

I only know that $2y\cdot \frac {dy}{dx}=2x (a^2-2x^2)$ so when $\frac {dy}{dx}=0, x^2=a^2/2.$

Thanks.

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  • $\begingroup$ Have you noticed that this is an array of curves? $\endgroup$ – callculus Feb 9 at 15:14
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Note that there is a symmetry about $x-$axis and also about $y-$axis. Thus you can reduce your sketch to the first quadrant. Polar coordinates will help.

Wolframalpha gives this for $a=1, a=2, a=4.$ enter image description here

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Rewriting $y^2=x^2(a^2-x^2)$ using the substitution $x = a \cos \theta$, we get its parametric form:

$( a \cos \theta,a^2 \cos \theta \sin \theta ), \ \theta \in [-\pi,\pi)$

which is a family of Lissajous-like curves, one of which for $a=1$ is plotted below

enter image description here

Plots of remaining ones for $-10 \le a \le 10$ can be explored here

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