The question asks:
Parametrize the intersection of the surfaces using trigonometric functions. $$y^2-z^2=x-6$$ $$y^2+z^2=81$$
$\mathbf{r}(t)=$ ____
My first step was recognizing $y^2+z^2=81$ as a circle in the $yz$-plane, which could be defined as $y=9\cos(t)$ and $z=9\sin(t)$.
I then thought that I could substitute in the above values of $y$ and $z$ into the equation $\mathbf{r}(t)=\boxed{\langle9\cos(t)-9\sin(t)+6,9\cos(t),9\sin(t)\rangle}$, but apparently that is incorrect.
Any advice to get me untangled and back on the correct path would be much appreciated.
$ x = y^2 - z^2 + 6 = 81(\cos^2 t - \sin^2 t) + 6 $
The parametrization is $ (81(\cos^2 t - \sin^2 t) + 6, 9\cos t, 9\sin t) $