Answer
$(2,\frac{1}{2},-1)$
and
$(-2,-\frac{1}{2},1)$
Work Step by Step
The equation for the hyperboloid is:
$f(x,y,z)=x^2+4y^2-z^2-4=0$ ... (1)
Normal to a tangent plane to the surface is $(2x,8y,-2z)$. This vector is parallel to the vector $(2,2,1)$ if and only if:
$2x=2k$, $8y=2k$ and $-2z=k$
Thus, we have
$x=k$, $y=k/4$ and $z=-k/2$
Equation (1) becomes:
$k^2+4(k/4)^2-(-k/2)^2-4=0$
$k^2=4$ or $k = \pm 2$
For $k=2$, we have points: $(2,\frac{1}{2},-1)$
For $k=-2$, we have points: $(-2,-\frac{1}{2},1)$
