The problem is:

"Parametrize the intersection of the hyperboloid $\displaystyle x^2+y^2=z^2+1 $ and the plane $\displaystyle z=ay $, where $\displaystyle a$ is a positive real number. Give the range of the values taken by the parameter. -Hint: Consider the cases 0<a<1 and a > 1 separately, and remember that x, y and z are all real."

So here's what I did:

I first plugged in the equation of the plane into the equation of the hyperboloid: $\displaystyle x^2+(1-a^2)y^2=1 $

If a<1, I get an ellipse of the form: $\displaystyle \frac {x^2}{1^2} +\frac{y^2}{\frac{1}{(\sqrt{1-a^2})^2}}=1$, similar to $\displaystyle \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$

From an online source, I found that $\displaystyle x=\cos\phi$ and $\displaystyle y=\frac{1}{\sqrt{1-a^2}}\sin\phi $, and therefore $\displaystyle z=\frac{a}{\sqrt{1-a^2}}\sin\phi$ and that $\displaystyle 0\leq \phi \leq 2\pi $

If a>1, the equation can be put in the form $\displaystyle \frac {x^2}{1^2} -\frac{y^2}{\frac{1}{(\sqrt{a^2-1})^2}}=1$ which is the equation of a hyperbola. Once again, from an online source (Wolfram MathWorld), I found that I could represent both branches of the hyperbola with $\displaystyle x=\sec\phi$, $\displaystyle y=\frac{1}{\sqrt{a^2-1}}\tan\phi $ and $\displaystyle z=\frac{a}{\sqrt{a^2-1}}\tan\phi$, where $\displaystyle -\pi < \phi < \pi $ with discontinuities at $\displaystyle \pm \frac{\pi}{2} $

I've seriously spent a lot of time on this problem, and this is the best I can come up with. Can someone tell me if it's acceptable? If not, what should I do?