Do the graphs of $r=\sec 2\theta$ and $r=\csc 2\theta$ consist of two hyperbolas each?

Question

The graphs of the polar equations $r=\sec 2\theta$ and $r=\csc 2\theta$ look like the multiplication and addition signs respectively. They also appear to consist of two hyperbolas each. Is that in fact the case?

If you enter the equations on Desmos, you would get the following:

The red curve is the graph of $r=\sec 2\theta$ and the blue one is the graph of $r=\csc 2\theta$.

Answer 1

I'll transform $r=\sec{(2\theta)}$ to Cartesian coordinates, and you can do the same for $r=\csc{(2\theta)}$.

$$r=\sec{(2\theta)}$$

$$r=\frac{1}{\cos{(2\theta)}}$$

$$r=\frac{1}{\cos^2(\theta)-\sin^2(\theta)}$$

$$\sqrt{x^2+y^2}=\frac{1}{\cos^2(\arctan(\frac{y}{x}))-\sin^2(\arctan(\frac{y}{x}))}$$

$$\sqrt{x^2+y^2} = \frac{1}{\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2 - \left(\frac{y}{\sqrt{x^2+y^2}}\right)^2}$$

$$\sqrt{x^2+y^2}=\frac{x^2+y^2}{x^2-y^2}$$ $$x^2-y^2=\sqrt{x^2+y^2}$$

So.... not exactly a hyperbola. This last line will give you one of the "hyperbolas", and if you do $-\sqrt{x^2+y^2}$ instead, you'll get the other "hyperbola". So... the actuall equation is $x^2-y^2=\pm \sqrt{x^2+y^2}$. Or, square both sides and get $x^2+y^2=x^4-2x^2y^2+y^4$. Hope this helps.

Answer 2

Take the curve $r = \sec(2\theta)$.

We have $r \cos(2 \theta)=1$, then $r (\cos^2(\theta) - \sin^2(\theta)) = 1$.

Multiply by $r$ and use $x = r \cos(\theta)$ and $y = r\sin(\theta)$ to get $x^2-y^2=r$.

Squaring and using $r^2 = x^2+y^2$ we get the cartesian equation of the curve:

$$(x^2-y^2)^2 = x^2+y^2$$

This is a quartic, not a quadratic equation, so it's not a hyperbola.