The figure below shows a conic section (specifically an ellipse) with focus $F$ and directrix $D$. The conic is the intersection of the blue cone and the purple "cutting plane". The figure also shows the Dandelin sphere associated with $F$ and $D$. (Interestingly, the Wikipedia article mentions the focus-directrix property "can be proved" with Dandelin spheres, but doesn't give that proof!) Let's call the green "horizontal" plane containing the circle where the sphere meets the cone the Dandelin plane; call that circle the Dandelin circle.
Now, given $P$ on the conic, let $Q$ be the corresponding point on the Dandelin circle; that is, let $Q$ be the point where the segment joining $P$ to the cone apex meets the Dandelin plane. Let $R$ be the foot of the perpendicular dropped from $P$ to the Dandelin plane, and let $S$ be the foot of the perpendicular dropped from $P$ to the directrix.
(Original image credit, with description of Dandelin spheres.)
Since segments $\overline{PF}$ and $\overline{PQ}$ are both tangent to the Dandelin sphere, we must have $|\overline{PF}|=|\overline{PQ}|$. (This, by the way, is the primary magic of the Dandelin sphere.)
We can massage the focus-directrix ratio for $P$ thusly:
$$\frac{|\overline{PF}|}{|\overline{PS}|} = \frac{|\overline{PQ}|}{|\overline{PS}|}=\frac{|\overline{PR}|/(\sin\angle Q)}{|\overline{PS}|}=\frac{|\overline{PR}|}{|\overline{PS}|}\frac{1}{\sin\angle Q}=\frac{\sin \angle S}{\sin \angle Q} \tag{$\star$}$$
Clearly, $\angle S$ is constant as $P$ moves about the conic; it's the angle between the cutting plane and Dandelin plane. But $\angle Q$ is also constant: it's the ("exterior") angle that the surface (more precisely, a "generator" line) of the cone makes with the Dandelin plane. Therefore, the focus-directrix ratio is a constant.
To complete the answer to your question, all we have to do is prove that "(focal distance)-over-(major radius)" gives the same trigonometric ratio.
For now, we'll assume the conic is an ellipse (that is $\angle S$ is smaller than $\angle Q$).
Look at the figure "sideways", reducing all the elements to their intersections with the plane through the cone's axis, perpendicular to the directrix. I'll take $P$ to be the point on the conic closest to the directrix (which itself has projected into the point $S$), and $P^\prime$ the farthest point. ($Q$ and $Q^\prime$ are the corresponding points on the Dandelin plane, which has projected into a line.) Then $\overline{PP^\prime}$ is the major axis of the ellipse. The ellipse's focus, $F$, corresponds to the point where the incircle of $\triangle OPP^\prime$ meets $\overline{PP^\prime}$; the ellipse's center corresponds to $M$, the midpoint of $\overline{PP^\prime}$.
Now that we know where everything is, a couple more applications of the equal-tangent-segment property are all we need. With $a := |\overline{MP^\prime}|$ and $c := |\overline{MF}|$, we have
$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| - |\overline{QP}|}{2a} = \frac{(a+c)-(a-c)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star$}$$
For the hyperbola, overlapping elements muddle the diagram a bit, but the argument is essentially the same (with a simple sign change):
$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| + |\overline{QP}|}{2a} = \frac{(a+c)+(c-a)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star^\prime$}$$
There's no argument to make for the parabola, which has no "focal distance" or "major radius". However, one sees that, as the $\angle S$ nears $\angle Q$, the ratio of the lengths of these elements within an ellipse or hyperbola approaches $1$, as expected.
Thus, the "distance-to-focus-over-distance-to-directrix" ratio and the "focal-radius-over-major-radius" ratio (when defined) are the same constant that we happen to call the "eccentricity" of a conic. This discussion reveals the geometric meaning of that number. I suspect that most students these days had no idea that there is such meaning. Kudos to the teacher who assigned this problem as homework.
One can use a list of equations to determine the property you require. Note, however, that in many cases it is easier to use derived value(s) as opposed to using equations that rely solely on the coefficients of the equation in general quadratic form. Below are examples of equations one can use.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form:
$ a x^2 + b x y + c y^2 + c x + e y + f = 0 $
The equation represents an ellipse if:
$ b^2 - 4 a c < 0 $
or similarly,
$ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The latus rectum is given by:
$ l = 2 {{ {r_\min}^2 } \over {r_\max}} $
The eccentricity is given by:
$ g = {{s} \over {r_\max}} $
The distance between center and closest directix point (either of the two) is given by:
$ h = {{{r_\max}^2} \over {s}} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
The focal points are given by:
$ F_{1,x} = x_\Delta - s \ cos (\theta) $
$ F_{1,y} = y_\Delta - s \ sin (\theta)) $
$ F_{2,x} = x_\Delta + s \ cos (\theta) $
$ F_{2,y} = y_\Delta + s \ sin (\theta)) $
(I tried to enter the equations without error. If you find an error, please post a comment)
Best Answer
The appropriate setting for this is the complex projective plane. While declaring some symbols, I will talk a tiny bit about that, but do not mistake this as a proper introduction to the subject.
In the projective plane, we use triples $(X:Y:Z)$ of homogenous coordinates for points, not all equal to zero. Two such triples are considered equal if one is a scalar multiple of the other. Affine points $(x,y)$ are embedded as $(x:y:1)$. The equation for the conic in homogenous coordinates is then $$\begin{pmatrix}X\\Y\\Z\end{pmatrix}^\top \underbrace{ \begin{pmatrix}2A & B & D\\B & 2C & E\\D & E & 2F\end{pmatrix} }_M \begin{pmatrix}X\\Y\\Z\end{pmatrix} = 0\tag{1}$$ with the matrix $M$ being regular for a non-degenerate conic.
We also use triples $[U:V:W]$ to represent lines, with a point $(X:Y:Z)$ being on a line $[U:V:W]$ if $UX+VY+WZ = 0$. The equation for the tangents to the conic then takes the form $$\begin{pmatrix}U\\V\\W\end{pmatrix}^\top \underbrace{ \begin{pmatrix}2G & H & R\\H & 2K & S\\R & S & 2T\end{pmatrix} }_{L} \begin{pmatrix}U\\V\\W\end{pmatrix} = 0\tag{2}$$ where $L$ is the matrix inverse of $M$, multiplied with an arbitrary nonzero scalar. Therefore, when given $M$, we can compute a suitable $L$, and if we set $L$ to the adjugate matrix of $M$, we do not even need to carry out divisions for that.
Much useful information about the conic is easier to find in $L$ than in $M$. Examples:
By the way, we can interpret $M$ as a linear map from projective points to projective lines. The linear map represented by $L$ does the reverse. Every point and line thus related form a (pole,polar) pair with respect to the given conic. I will use the linear map feature later. Let me just mention here that the conic equations $(1)$ and $(2)$ can be interpreted as stating that the conic consists of those points that lie on their polars, and in that case the polars are tangents to the conic.
Finally, complex coordinates. We need those for the following:
Proposition 1: Let $F = (x_F:y_F:1)$ be a real finite focus of a non-degenerate conic section with real coefficients. Let $z_F = x_F + \mathrm{i}y_F$ where $\mathrm{i}$ is the imaginary unit. Then the conic has the complex tangent $$q_F = [1:\mathrm{i}:-z_F]\tag{3}$$ In other words, $(2)$ is fulfilled for $[U:V:W] = [1:\mathrm{i}:-z_F]$. The proof is an algebraic exercise left to those readers who can express the entries of the matrix $L$ in terms of focus, directrix, and excentricity from first principles. An expression for the elliptic case is given in another answer.
Note that the line $q_F$ passes through $F$: Just do the dot product to see that. You might take that as a hint at the reason why we need complex numbers here: There are no real tangents to a conic that pass through a focus or through any other interior point.
To apply proposition 1, we plug $[U:V:W] = [1:\mathrm{i}:-z_F]$ into $(2)$. This yields the following equation for $z_F$, with complex coefficients:
$$T\,z_F^2 - (R + \mathrm{i} S)\,z_F + (G - K + \mathrm{i} H) = 0\tag{4}$$
If $T=0$ (parabola), then $(4)$ has a unique solution for $z_F$, giving the finite focus of the parabola. If $T\neq 0$, then $(4)$ is a quadratic equation for $z_F$ whose solutions give both foci. In the case of a circle, both foci coincede.
So basically one just has to memorize "tangent $[1:\mathrm{i}:-z_F]$" in order to deduce equation $(4)$ for the foci.
Now, how to get the directrix associated with the focus $F$?
Proposition 2: The directrix is the polar of the focus.
Therefore, if the directrix is to be presented as a triple $d_F = [U_d:V_d:W_d]$ and the focus is given as $F = (x_F:y_F:1)$, then one just has to compute $$d_F = M\cdot F\tag{5}$$ The proof of that is again left as algebraic exercise. Use the directrix-based formulation of the conic equation to express $M$ suitably.