How to find the parametric equation of: $x^2-3y^2=1$

Question

Question

How to find the parametric equation of: $x^2-3y^2=1$

I know that $x=\cos{\theta}$ and $x=\sin{\theta}$

But I've come up with the bogus solution that the answer is:

$\cos{\theta}^2-3(\sin{\theta})^2=1$

I was wondering what was the real way in parametricizing this equation?

Answer 1

write $$x^2-3y^2=1$$ $$x^2-(\sqrt{3}y)^2=1$$ compare with $$\cosh^2t-\sinh^2t=1$$ so we have \begin{cases} x=\cosh t\\ \sqrt{3}y=\sinh t \end{cases} or \begin{cases} x=\cosh t\\ y=\dfrac{1}{\sqrt{3}}\sinh t \end{cases}

Answer 2

What about $$x=\sec t,\quad y=\frac{\tan t}{\sqrt{3}}\;\LARGE ?$$