How do I describe the curve $x = \sqrt{1 + y^{2}}$ in polar coordinates?

Question

Could someone explain to me how this conversion from cartesian to polar form is done with this type of equation?

Show that the curve $x = \sqrt{1+y^2}$ can be described in polar coordinates by $$r^2 = \frac 1 {\cos^2 \theta - \sin^2 \theta}.$$

Answer 1

Assuming that $x \geq 1$, one gets the following result:

\begin{align*} x = \sqrt{1 + y^{2}} & \Longleftrightarrow x^{2} = 1 + y^{2}\\\\ & \Longleftrightarrow x^{2} - y^{2} = 1\\\\ & \Longleftrightarrow r^{2}(\cos^{2}(\theta) - \sin^{2}(\theta)) = 1\\\\ & \Longleftrightarrow r^{2}\cos(2\theta) = 1 \end{align*}

Based on it, I think you meant \begin{align*} r^{2} = \frac{1}{\cos^{2}(\theta) \color{red}{-} \sin^{2}(\theta)} \end{align*}

Hopefully this helps!