The equation
$$\vec r = \vec r_0 + t \vec v$$
is a parametric equation for a line. You can see $\vec r$ as a function of $t$:
$$\vec r(t) = r_0 + tv$$
and for any specific $t_0$, $\vec r(t_0)$ will be a vector pointing to a point on the line.
The whole line is the set of all points which are produced by $\vec r(t)$, i.e. the set
$$\{\vec r(t) \mid t \in \mathbb R\}.$$
Thus, it does not make sense to ask "how to solve for $t$?" as your question is phrased.
However, if you have a point $\vec p$ and ask the question "for which $t$ is $\vec r(t) = \vec p$?", that does make sense. In this case you solve the equation
$$\vec p = \vec r(t) = \vec r_0 + t \vec v$$
and this will be solvable only when $\vec p$ is on the line.
Being given a general quadric with implicit equation:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gx+2hy+2iz+k=0,$$
the key point is that the tangent plane to this quadric in $(x',y',z')$ is obtained by "polarization", i.e., by looking at the homogeneous version of (1). This homogeneous form is obtained by introducing a 4th variable $t$, atop variables $x,y,z$, in such a way that if a term is only 1st degree, it is multiplied by $t$, and if it is a constant, it is multiplied by $t^2$: thus, all terms become of degree 2 (whence the name "homogenization"), in this way:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gtx+2hty+2itz+kt^2=0$$
Remark: if $t=1$, one finds back (1).
Then the method amounts to build the bilinear form associated with this quadratic form in 4 variables:
$\tag{2}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(t'x+x't)+h(t'y+ty')+i(t'z+tz')+ktt'=0.$
Remark: if all the "primes" are suppressed, we are back to equation (2).
Then make $t=t'=1$ in (2),
$$\tag{3}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(x+x')+h(y+y')+i(z+z')+k=0,$$
We have obtained in (3) the equation of the tangent plane to the quadric at point $(x',y',t')$.
Remark: this techniques can be justified in the framework of projective geometry.
Applying this to our case, we get, for the tangent plane at $(x_0,y_0,z_0)$ is
$$xx_0+4yy_0+zz_0=9$$
This is the equation of a plane with normal vector
$$\pmatrix{x_0\\4y_0\\z_0} \text{which is desired to be proportional to} \pmatrix{-4\\8\\-2}.$$
Thus, we must look for a coefficient $\alpha$ such that:
$$\tag{4}\pmatrix{x_0\\4y_0\\z_0}=\pmatrix{-4\alpha\\8\alpha\\-2\alpha}$$
Point $(x_0,y_0,z_0)$ is a point of the ellipsoid if and only if:
$$(-4\alpha)^2+4(2\alpha)^2+(-2\alpha)^2=9$$
i.e., $$\alpha^2=\frac{9}{36}=\frac{1}{4} \ \ \Leftrightarrow \ \ \alpha=\pm \frac{1}{2}$$
Whence the two solution points by plugging these opposite values of $\alpha$ in (4):
$$\pmatrix{x_0\\y_0\\z_0}=\pm \frac{1}{2} \pmatrix{-4\\2\\-2}=\pm \pmatrix{-2\\1\\-1}.$$
Best Answer
It is likely more helpful to set out the problem using vector forms for the relevant lines. The vector $ \ \vec{c} \ = \ \vec{AB} \ $ is $ \ \vec{c} \ = \ \vec{b} \ - \ \vec{a} \ $ (since $ \ \vec{a} \ + \ \vec{c} \ = \ \vec{b} \ $ ) . It appears that you are working in $ \ \mathbb{R}^2 \ $ , so we have
$$ \vec{a} \ = \ \langle x_1 \ , \ y_1 \rangle \ \ , \ \ \vec{b} \ = \ \langle x_2 \ , \ y_2 \rangle \ \ , \ \ \vec{c} \ = \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
It is convenient to use $ \ \vec{a} \ $ as the "initial" vector with a parameterization that places $ \ t = 0 \ $ there (point A) and places $ \ t = 1 \ $ at $ \ \vec{b} \ $ (point B) , with the direction given by $ \ \vec{c} \ $ . The vector $ \ \vec{c} \ $ then lies along the line AB having parametric equation
$$ \ \vec{r} \ = \ \vec{a} \ + \ t \ \vec{c} \ = \ \langle x_1 \ , \ y_1 \rangle \ + \ t \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
The midpoint is then located at $ \ t = \frac{1}{2} \ $ , said point being at the head of position vector $ \ \langle \frac{x_1 \ + \ x_2}{2} \ , $ $ \ \frac{y_1 \ + \ y_2}{2} \rangle \ $ .
The perpendicular bisector of $ \ \overline{AB} \ $ can be defined by a vector which has a zero "dot product" (scalar product) with $ \ \vec{c} \ $ . If we say that this line has a slope $ \ m \ $ , we can write its direction as $ \ \langle 1 \ , \ m \rangle \ $ , and so
$$ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \cdot \ \langle 1 \ , \ m \rangle \ \ = \ \ (x_2 - x_1 ) \ + \ m \ (y_2 - y_1) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ m \ \ = \ \ - \frac{x_2 - x_1}{y_2 - y_1} \ \ . $$
(Note that this is just what we expect from the "product of slopes of perpendicular lines" theorem, since the direction $ \ \vec{c} \ $ has slope $ \ \frac{y_2 - y_1}{x_2 - x_1} \ $ . )
Since $ \ y_2 - y_1 \ \neq \ 0 \ $ , we can then also write the direction for the perpendicular bisector as $ \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ $ . (It is also acceptable to attach the "minus sign" to the first component instead -- thus, $ \ \langle y_1 - y_2 \ , \ x_2 - x_1 \rangle \ $ -- which is a vector pointing in the opposite direction on the perpendicular bisector line.
Finally, we can establish a parametrization of the perpendicular bisector with the midpoint of $ \ \overline{AB} \ $ at $ \ \tau = 0 \ $ , making it the "initial" vector for the bisector. The vector form for this line may then be written as
$$ \vec{R} \ = \ \langle \frac{x_1 \ + \ x_2}{2} \ , \ \frac{y_1 \ + \ y_2}{2} \rangle \ + \ \tau \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ \ . $$
This method can be extended to $ \ \mathbb{R}^3 \ $ , except that the perpendicular bisector is ambiguous to the extent that it lies in a plane for which $ \ \vec{AB} \ $ is its normal. To pin things down, we would need to specify a point in that plane that the bisector passes through. (This is sometimes given as a textbook or exam problem.)