Finding direction vector Can someone please explain how the direction vector was found in problem $2$ of this worksheet?
Below is an image of the problem $2$ of the worksheet.

 A: Firstly by the observation direction vector should lie at the $zx$ plane so $y$ in the direction vector should be $0$. Secondly after that he is thinking of the slope of the direction vector as the direction vector of the fuction $ z(x)$ where the direction vector is that case is $(1,z'(x_0))$
A: You want to know the tangent line to the ellipse at the given point along the plane $y=2$, so go ahead and plug in $y=2$ to obtain  $4x^2 + 8 + z^2 = 16$. Take the total differential to obtain
$$ 8x dx  + 2 z dz = 0.  
$$ 
This means 
$$
2z dz = -8x dx, \mbox{ or } \dfrac{dz}{dx} = -\dfrac{8x}{2z} = -\dfrac{4x}{z}.    
$$
Since 
$$ 
\dfrac{dz}{dx}|_{(x,z)=(1,2)} = -\dfrac{4}{2} = -2,   
$$
the line tangent to the intersection of the ellipsoid and the plane at the point $(1,2,2)$ has the direction vector 
$$
(1,0,-2). $$ 
Note that the tangent line is parallel to the $y=2$ plane so the vector should not be changing in the $y$-direction. Moreover, any nonzero scalar multiple of $(1,0,-2)$ will also work.  
Thus the tangent line is 
$$ 
l(t) = (1+t, 2, 2-2t), 
$$ 
or you could also write it as 
$$
l(t) = (1,2,2)+ t(1,0,-2), 
$$
where $t$ varies over the reals.  
