<s>[INST] How can I derive the polar representation of Bernoulli's lemniscate, $r^2 = a^2\cos 2\theta$? If I start with the formula $(x^2+y^2)^2 -a(x^2-y^2) = 0$, 
Converting to polar coordinates gives 
$r^2 - a(r^2\cos^2 \theta - r^2 \sin^2 \theta) = 0$. 
Applying the identity $ \cos^2 \theta - \sin^2 \theta = \cos 2\theta$,
I find that
$r^2 - ar^2(\cos 2\theta) = 0$
How do I derive $r^2 = a^2\cos 2\theta$ from here? 
 [/INST] You forgot to square the first term
$$
x^2+y^2 = r^2\implies (x^2+y^2)^2 = (r^2)^2 = r^4
$$
</s>