Calculus 3 Problem deals with partials A particle travels along the curve that lies on both the hyperboloid
$x^2 + y^2 − z^2 = 1$ and the plane $x + y + z = 0$. When it reaches the
point $(\frac23, −\frac34, \frac1{12})$ it flies off on a tangent, and thenceforth travels
along a straight line, until it hits the paraboloid $99z = 64y^2 − 81x^2$
.
Where does the particle hit the paraboloid? 
So i thought to solve the plane for each variable then plug them in to the hyperboloid. However im not even sure were to start please help...
 A: Hint: the direction vector that the particle flies off in should be perpendicular to both the normal vector of $x^2+y^2-z^2=1$ at $\left(\frac{2}{3}, - \frac{3}{4}, \frac{1}{12} \right)$ and the normal vector of $x+y+z=0$ at $\left(\frac{2}{3}, - \frac{3}{4}, \frac{1}{12} \right)$.
A: Since $z=-(x+y)$ from the plane, we can substitute into the hyperboloid equation to get
$$x^2 + y^2 - (-x-y)^2 = 1$$
$$x^2 + y^2 - x^2 -y^2 -2xy = 1$$
$$xy = -1/2$$
$$y = \frac{-1}{2x}$$
So the set of $(x,y,z) \in \mathbb{R}^3$ that are in this intersection of plane and hyperboloid is a curve parametrized by $$x(t) = t$$
$$y(t)  = \frac{-1}{2t}$$
$$z(t) = \frac{1}{2t}-t=\frac{1-2t^2}{2t}$$
As a check, make sure that $(2/3, -3/4, 1/12)$ is on this curve (it is). 

We have a vector function from $\mathbb{R}$ to $\mathbb{R}^3$: $$f(t) = (x(t), y(t), z(t)) = \left(t, \frac{-1}{2t}, \frac{1-2t^2}{2t}\right)$$
Now we can find $f'(t)$:
$$f'(t) = \left(1, \frac{1}{2t^2},-\frac{2t^2+1}{2t^2}\right)$$
The $t$ for which $f(t) = (2/3, -3/4, 1/12)$ is simply the $x$-coordinate, or $t=2/3$. So we need to find $f'(2/3)$:
$$f'\left(\frac{2}{3}\right) = \left(1,\frac{9}{8}, -\frac{17}{8}\right)$$

So we have the particle following a line with direction vector $f'(2/3)$, that goes through the point $f(2/3)$. 
You should be able to find an equation for this line. You will then be able to find where it intersects the paraboloid. 
I hope this helps :)  
