$$ r^2=\sec 4\theta $$
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!!!
$\endgroup$ 32 Answers
$\begingroup$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 $$\Longrightarrow x^4+y^4-6x^2y^2=x^2+y^2$$
$\endgroup$ 2 $\begingroup$i can take $$r^2 =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}$$ and turn it into a cartesian equation.
$\begin{align} 1 &=\frac{r^2}{r^4-8r^4\sin^2\theta + 8r^4\sin^4\theta}\\ &=\dfrac{(x^2+y^2)}{(x^2+y^2)^2 - 8(x^2+y^2)y^2 +8y^4}\\ &= \dfrac{(x^2+y^2)}{x^4 + y^4-6x^2y^2}\\ \end{align}$
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