Which of the two $k$ values do I use in solving for $r'$ and $z'$ ?
Either one will be a point of intersection, since in general a line intersects an ellipse in two points. Which one you need, or if it doesn't matter, depends on your application.
How can you tell when the line doesn't intersect the ellipse? Is the quadratic equation for $k$ not have any real roots?
Right. And if you only have a single root, i.e. both roots coincide, then the line will be a tangent.
Is it important for $r$ to be $>0$ in for this equation even though I'm not testing a point, but a line instead?
I see no reason to require this here.
Why is the semi-minor axis being defined as $a_e(1-f)$? I usually just define the semi-minor axis the same way I define the major, so could I just replace all the $(1-f)$s in the above equations with my desired semi-minor axis length?
Yes, you can write the semi-minor axis (called $a_p$ in your document) instead of $a_e(1-f)$. Which means you could replace the $(1-f)$ themselves by $\frac{a_p}{a_e}$ or multiply the whole equation by $a_e^2$.
Finally, is there any simpler, faster way to see if and were an ellipse and line intersect?
Depends on how your objects are given. If you really have an ellipse and a line in the described form, I can't think of something fundamentally easier. If, on the other hand, you have an ellipse in some other representation, then you might be able to use that representation directly to compute the intersection, instead of transforming its representation first. In any case, you can't avoid the quadratic equation.
What I'd do
I read up to $\phi=\psi-\alpha$ and agree with that. I'm spinning my own thoughts from that point on. For the moment I'd like to think of your tangent direction not as an angle but as a direction vector instead, namely the vector
$$ v_1 = \begin{pmatrix} \cos\phi \\ \sin\phi \end{pmatrix} $$
Now you can take your whole scene and scale all $x$ coordinates by $a$ and all $y$ coordinates by $b$. This will turn your ellipsis into a unit circle, and the tangent direction will become
$$ v_2 = \begin{pmatrix} \frac{\cos\phi}a \\ \frac{\sin\phi}b \end{pmatrix} $$
Now you're looking for the point on the unit circle with this tangent. This is particularly easy, since tangents to the unit circle are perpendicular to radii. Simply take your vector, swap $x$ and $y$ coordinate and also swap one sign. That will result in a perpendicular vector
$$ v_3 = \begin{pmatrix} \frac{\sin\phi}b \\ -\frac{\cos\phi}a \end{pmatrix} $$
Now you have to change the length of that vector to $1$ so you get a point on the unit circle.
$$ v_4 = \frac{v_3}{\lVert v_3\rVert} =
\frac{1}{\sqrt{\left(\frac{\sin\phi}b\right)^2 + \left(\frac{\cos\phi}a\right)^2}}
\begin{pmatrix} \frac{\sin\phi}b \\ -\frac{\cos\phi}a \end{pmatrix}
= \begin{pmatrix} x_4 \\ y_4 \end{pmatrix} $$
Note that at this point, the opposite point $-v_4$ is a second valid solution.
Now you can scale your coordinates back by $a$ and $b$ and end up with
$$ v_5 = \begin{pmatrix} a\cdot x_4 \\ b\cdot y_4 \end{pmatrix} $$
Last you'd apply the rotation by $\alpha$ to that (and possibly $-v_5$ as well).
What you did
So now that I've thought about how I'd think about this, I'll have a look at the rest of what you did. It seems that your computations look a lot shorter than mine, so they might be more efficient for practical uses. Nevertheless, my approach might yield some insight as to what the individual steps do, so I'll leave it in place and even refer to it.
I found that I had to move that $−$ from the $y$ part to the $x$ part otherwise my calculation was off by $180°$
If your tangents are unoriented lines, then a change in $180°$ in that argument will give an equally valid result. This is the $v_4$/$-v_4$ ambiguity in my solution. The “velocity vector” you used is oriented, pointing in a given direction, but if you move along your circle in the opposite direction, you'd get opposite velocities at the same points.
So again the question: how does the above solution work, or what is a better more proper rigorous derivation of a solution to the problem?
Your solution looks good. You might want to consider $t$ and $t+180°$, and if you do, then it shouldn't matter where you place your minus sign. If you still have doubts, however, feel free to implement my approach as an alternative and compare the results. They should agree.
I should also note that http://mathworld.wolfram.com/Ellipse.html eqn 60 gives the relation between $\phi$ and $t$ as […]
Their $\phi$ is the “polar angle” of a point on the ellipsis. Look at the line connecting the center of the ellipsis with a given point on the ellipsis. The angle that line makes with the $x$ axis is their $\phi$.
Best Answer
Draw points like this, as many as you wish: