In principle it is quite simple to find what you need, one only needs to discuss two different cases: whether the hyperboloid is of one or two sheets. I'll treat the one sheet case, the other is completely analogous.
The equation of a one-sheet hyperboloid, symmetric around the $x$ axis is:
$$
{y^2+z^2\over a^2}-{x^2\over b^2}=1.
$$
Intersecting that with the $(x,y)$ plane (equation $z=0$) gives the equation of a hyperbola of equation $y^2/a^2-x^2/b^2=1$, with semi-major axis $a$ and semi-minor axis $b$.
Intersecting instead with the a plane parallel to the $(x,y)$ plane and at a distance $d$ from it (equation $z=d$) gives the equation of a hyperbola of equation $y^2/a^2-x^2/b^2=1-d^2/a^2$, that is:
$$
{y^2\over\beta a^2}-{x^2\over\beta b^2}=1,
$$
where $\beta=1-d^2/a^2$.
If $\beta>0$ (that is $d<a$) this is the equation of a hyperbola intersecting the $y$ axis, with semi-major axis $\sqrt\beta a$ and semi-minor axis $\sqrt\beta b$.
If $\beta<0$ ($d>a$) this is the equation of a hyperbola intersecting the $x$ axis, with semi-major axis $\sqrt{-\beta} b$ and semi-minor axis $\sqrt{-\beta} a$.
If $\beta=0$ ($d=a$) the curve degenerates into a pair of straight lines of equations $y=\pm (a/b)x$, which are also the common asymptotes of all the hyperbolas obtained for $\beta\ne0$.
Notice that the eccentricity of these hyperbolas is fixed:
$$
e={\sqrt{|\beta| a^2+|\beta| b^2}\over \sqrt{|\beta|} a}={\sqrt{a^2+ b^2}\over a},
$$
while the semi-latus rectum gets multiplied by $\sqrt{|\beta|}$:
$$
L={(\sqrt{|\beta|} b)^2\over\sqrt{|\beta|} a}=\sqrt{|\beta|} {b^2\over a}
\ \ \hbox{for $\beta>0$, while}\ \ L=\sqrt{|\beta|} {a^2\over b}
\ \ \hbox{for $\beta<0$.}
$$
Foci lie on the $y$ axis if $\beta>0$, and on the $x$ axis if $\beta<0$, at a distance $\sqrt{|\beta|}\sqrt{a^2+b^2}$ from the origin.
Let the point of tangency be $(x_1,y_1)$
Then the equation of tangent is $\displaystyle\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$.
When $y=0$, $\displaystyle x=x_1+\frac{-y_1}{3x_1^2-6x_1-7}$.
When $x=0$, $\displaystyle y=y_1-x_1(3x_1^2-6x_1-7)$.
\begin{align}
-2\left(x_1+\frac{-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\
-2\left(\frac{x_1(3x_1^2-6x_1-7)-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\
2&=3x_1^2-6x_1-7\\
3x_1^2-6x_1-9&=0\\
x_1&=-1\quad\text{or}\quad3
\end{align}
When $x_1=-1$, $y_1=(-1)^3-3(-1)^2-7(-1)+6=9$ and the $y$-intercept is
$$9-(-1)[3(-1)^2-6(-1)-7]=11>0$$
When $x_1=3$, $y_1=(3)^3-3(3)^2-7(3)+6=-15$ and the $y$-intercept is
$$-15-(3)[3(3)^2-6(3)-7]=-21<0$$
So, the point of tangency is $(-1,9)$.
Best Answer
Plug $z=\cos t, x=\sin t$ into the first equation: $$\sin^2t+y^2+4\cos^2t=4$$ We get $y^2=3-3\cos^2t=3\sin^2t$, since $y>0$, $y=\sqrt{3}|\sin t|$.