# 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!!!

• look at the graph of $r^ = sec 4\theta$. looks like there $8$ branches of hyperbola. so cartesian equation is bound to be complicated. – abel Jan 18 '15 at 1:29
• I solved this while the others had posted their answers and I came to the conclusion that $$r^2 = \frac{1}{8\cos^4 \theta - 8\cos^2 \theta + 1}$$ using double angle rules for $\cos (x)$. – Mohamad Ali Baydoun Jan 18 '15 at 1:43
• @ Mohammad Ali Baydoun I believe you can get that from using the identity $\cos2\theta = 2\cos^2\theta - 1$ where I had used $\cos2\theta = 1-2\sin^2\theta$ – mathamphetamines Jan 18 '15 at 1:44

Hint $$\dfrac{y^2}{x^2}=\tan^2{\theta}=\dfrac{\sin^2{\theta}}{1-\sin^2{\theta}}\Longrightarrow \sin^2{\theta}=\dfrac{y^2}{x^2+y^2}$$ and you have $$x^2+y^2=r^2=\dfrac{1}{1-8\sin^2{\theta}+8\sin^4{\theta}}=\dfrac{1}{1-\dfrac{8y^2}{x^2+y^2}+8\left(\dfrac{y^2}{x^2+y^2}\right)^2}$$ so $$x^2+y^2=\dfrac{(x^2+y^2)^2}{(x^2+y^2)^2-8y^2(x^2+y^2)+8y^4}$$ so $$\Longrightarrow x^4+y^4-6x^2y^2=x^2+y^2$$
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}