$$
r^2=\sec 4\theta
$$

I graphed this equations using Wolfram Alpha and found it to be 2 hyperbolas. I'm having difficulty showing this using the standard equations

$$
x=r\cos\theta \;, \; y=r\sin\theta \;, and \; x^2 +y^2 =r^2
$$

My work so far:

$$
r^2 = \sec4\theta=\frac{1}{\cos4\theta}=\frac{1}{\cos(2\theta+2\theta)}=\frac{1}{\cos2\theta \cos2\theta - \sin2\theta\sin2\theta} \\ \\ =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}
$$

I'm getting nowhere from here. I've tried using a few other trig identities, but no luck! Can one please point me in the right direction? I would appreciate any help. Thank you!!!