Tangent Plane Question. Find the tangent plane to the surface
$$4x^2 + y^2 - z^2 = 4$$
at the point $(1,-2, 2)$. 
Sketch the level curves for $z = k$ and for $y = k$ where
$k = -1, 0, 1, 2$. Hence sketch the surface.
Find $\theta$ such that
$r(t) = (\cos(\theta) + t \sin(\theta), 2 \sin(\theta) - 2t \cos(\theta), 2t)$
is the intersection of the surface with the tangent plane at $(1,-2, 2)$. 
Describe $r(t)$ and find the unit tangent vector to $r(t)$ at $(1,-2, 2)$.
what i've done so far
I've worked the tangent plane to be $z=2x-y-2$ 
but im not too sure how to draw the level curves or find $\theta$
 A: I would take another approach. Hope it helps.
Take $$G(x,y,z)=4x^2+y^2-z^2-4$$
The gradient of that function at the point is normal to the level curves (in this case we need $G(x,y,z)=0$)
$$\nabla G(x,y,z)=(8x,2y,-2z)$$
Evaluated at that point:
$$\nabla G(1,-2,2)=(8,-4,-4)$$
Then with the normal and a point we have the equation of the plane:
$$((x,y,z)-(1,-2,2))\cdot (8,-4,-4)=0$$
Which is:
$$z=2x-y-2$$
If $z=k$ then:
$$4x^2+y^2=4+k^2$$ 
Then dividing:
$$\frac{4}{4+k^2}x^2+\frac{1}{4+k^2}y^2=1$$ 
Which is a ellipse in the plane $z=k$
If $y=k$ then:
$$4x^2-z^2=4-k^2$$ 
Then dividing:
$$\frac{4}{4-k^2}x^2-\frac{1}{4-k^2}z^2=1$$ 
Which is a hyperbola in the plane $y=k$ according with the sign of $4-k^2$. 
But there is a case if $4=k^2$ then $z^2=4x^2$ which implies $z=-2x$ or $z=2x$ which are two lines in the plane $y=k$.
By definition $r(t)$ intersects the surface and the tangent plane at the point $(1,-2,2)$
$$(\cos(\theta) + t_0 \sin(\theta), 2 \sin(\theta) - 2t_0 \cos(\theta), 2t_0)=(1,-2,2)$$
Then:
$$\cos(\theta) + t_0 \sin(\theta)=1$$
$$2 \sin(\theta) - 2t_0 \cos(\theta)=-2$$
$$2t_0=2$$
Then $t_0=1$, replacing it:
$$\cos(\theta) + \sin(\theta)=1$$
$$2 \sin(\theta) - 2 \cos(\theta)=-2$$
Solving $cos(\theta)=1$ $sin(\theta)=0$. For example $\theta=0$ satisfy the condition but what do you really need is the value of $cos$ and $sin$.
Finally (replacing the value of $cos(\theta)$ and $sin(\theta)$):
$$r(t)=(1 , - 2t, 2t)$$
Then
$$r(t)=(1,0,0)+t(0,-2,2)$$
Which is a line with position at $(1,0,0)$ and direction $(0,-2,2)$. 
The tangent of a curve in a point is the derivative of the curve with respect at the variable:
$$r'(t)=(0,-2,2)$$
And the unitary tangent vector is the tangent vector normalized:
$$\hat r'(t)=r'(t)/||r'(t)||$$
Which is in this case is:
$$\hat r'(t)=(0,-2,2)/\sqrt{0^2+(-2)^2+2^2}=\frac{1}{\sqrt{8}}(0,-2,2)$$
