# Sketching curve

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.

• Have you noticed that this is an array of curves? Feb 9, 2019 at 15:14

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 $$a=1, a=2, a=4.$$
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
Plots of remaining ones for $$-10 \le a \le 10$$ can be explored here