Convert the Polar Equation to Cartesian Coordinates $$
r^2=\sec 4\theta
$$
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!!!

i can take $$r^2 =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}$$ and turn it into a cartesian equation.
$\begin{align}
1 &=\frac{r^2}{r^4-8r^4\sin^2\theta + 8r^4\sin^4\theta}\\
  &=\dfrac{(x^2+y^2)}{(x^2+y^2)^2 - 8(x^2+y^2)y^2 +8y^4}\\
  &= \dfrac{(x^2+y^2)}{x^4 + y^4-6x^2y^2}\\
\end{align}$
