Here's a better generalization than in my previous answer, using a different interpretation of that answer's equation $(3)$:
$$\left(\;y - \tfrac12\sin^2\theta\;\right) = -\frac{1}{2p\cos^2\theta}\left(\;x-\tfrac12\sin 2\theta\;\right)^2 \tag{1}$$
This time around, instead of interpreting $p\cos^2\theta$ as the semi-latus-rectum, we recognize it as twice the vertex-to-focus distance. This places the focus at
$$(\hat{x},\hat{y}) = \left(\;\tfrac12 p\sin 2\theta, \tfrac12 p\sin^2\theta - \tfrac12p\cos^2\theta\;\right) = \left(\;\tfrac12 p\sin 2\theta,-\tfrac12p\cos 2\theta\;\right) \tag{2}$$
which is on a circle of radius $r := \tfrac12p$. Let's use that as an alternative "third condition" from before; since the distance from center to focus is $ae$, we can replace the earlier equation $(6a)$ with
$$h^2 + ( k + ae )^2 = r^2 \tag{3}$$
Solving the new system for $a$, $h$, $k$ gives (upon discarding a couple of extraneous cases)
$$
a = \frac{\sigma r m (m + e \cos\theta)}{1 - e^2} \qquad
h = \sigma r \sin\theta (m + e \cos\theta) \qquad
k = - \frac{\sigma r \cos\theta (m + e \cos\theta)}{1 - e^2} \tag{4}$$
where $m := \sqrt{1-e^2\sin^2\theta}$ and $\sigma:=\pm 1$, so that the generalized conic has this equation:
$$x^2 + y^2 (1- e^2) + 2 \sigma r ( x \sin\theta - y \cos\theta ) \left( m + e \cos\theta \right) = 0 \tag{5}$$
The foci have coordinates
$$(\hat{x}, \hat{y}) = (h, k + \sigma_1 a) = \left(\; \sigma r\sin\theta(m+e\cos\theta), \;\frac{\sigma r (m + e \cos\theta) (\cos\theta -\sigma_1 m)}{1 - e^2}\;\right) \tag{6}$$
that satisfy
$$x^2 + y^2( 1 + \sigma_1 e)^2 + 2 \sigma \sigma_1 y r ( 1 + \sigma_1 e ) = 0 \tag{7}$$
That is,
$$\frac{x^2}{r^2} + \frac{(1 + \sigma_1 e)^2}{r^2} \left(y + \frac{\sigma\sigma_1 r}{1 + \sigma_1 e}\right)^2 = 1 \tag{8}$$
which indicates that each vertex of our conic has an elliptical locus. For $\sigma=1$, the horizontal axis is major; for $\sigma = -1$, that axis is minor. The corresponding eccentricities are
$$\sigma_1 = 1 \;:\;\frac{\sqrt{e(2+e)}}{1+e} \qquad \sigma_1 = -1\;:\; \sqrt{e(2-e)} \tag{9}$$
that, as with OP's original scenario, are independent of $r$ (or $p$, or $v$ and $g$). For $e=1$, these are $\sqrt{3}/4$ and $1$; the first confirms OP's observation, while the second is extraneous here.) $\square$
Here are some animated illustrations, showing scenarios for families of conics with eccentricity $0$, $0.5$, $0.999$, and $1.5$. (In the last case, observe that no drawing occurs while $1-e^2\sin^2\theta$ is negative.)
The reason that we can decompose the force $mg$ into components $mg\sin(30^\circ)$ and $mg\cos(30^\circ)$ is not because the concept of force has distance embedded in it (through the distance/time$^2$ in $g$). Rather it is simply because force is a vector quantity. Vectors arise often in physics, and it is important to make sense of them.
Before we can discuss vectors, we have to discuss coordinate systems. This is a way of defining a set of numbers (coordinates) to the system so that every point can be cataloged, so to speak, according to this system. Many systems (like the one you describe here) use a Cartesian coordiante system which consists of two (or more) axes which are perpendicular to each other, and each point is referenced according to its distance from each of these axes. We usually label the axes $x$ and $y$. We are used to $y$ being vertical and $x$ being horizontal. But this doesn't have to be the case. Depending on the problem you are trying to solve, it may make sense to rotate the axes. For instance, in your problem, we could put the $x$ axis in the plane of the ramp, with the $y$ axis perpendicular to it.
Now vectors. Vectors can be represented by a coordinate in the coordinate system (this is actually not generally true, but it will be in introductory physics, so will stick to that). For instance, a vector might have the representation $A=(3,4)$, which can be visualized as an arrow starting at the point $(0,0)$ and going out to the point $(3,4)$. We can also say that this vector is $A=A_x+A_y$ where $A_x=(3,0)$ and $A_y=(0,4)$. This is helpful because $A_x$ and $A_y$ are in the direction of the coordinate axes. Here, $A_x$ and $A_y$ are called components of $A$. But it is not always so easy to find the components. Suppose I said "$A$ is the vector which has a magnitude of $r=5$ and makes an angle of $\theta=36.9^\circ$ above the horizontal axis. This means that $A$ is represented by an arrow of length $5$ at this particular angle pointing slightly up above the horizontal. To find the components, we use trigonometry. By drawing a triangle like the one in the figure below, we find that the components are $A_x=(r\cos\theta,0)=(4,0)$ and $A_y=(0,r\sin\theta)=(0,3)$.
Note that I never had to assume that $A$ had anything to do with physical length here.
So let's go back to your problem. The force of gravity is a vector, and you want to compare it to the force of friction and acceleration (in the plane of the ramp), and the normal force (perpendicular to the ramp). This is a problem because these forces and acceleration are in line with a coordinate axis, but the force of gravity is not. To fix this, we just use the same trigonometry trick to break gravity up into its components: $F_g=(F\cos\theta,0)+(0,F\sin\theta).$
One final note. You may think it odd that I put a second coordinate which is just zero in each of the components (i.e., I wrote $(F\cos\theta,0)$ instead of just $F\cos\theta)$. This is just me being formal. The great thing about breaking every vector up into its components is that no you can compare all the components in the $x$ direction without worrying about those in the $y$ direction, and vice-versa. This is why components are useful.
In summary, breaking up a force into different pieces has to do with the fact that it is a vector, and not physical distance being part of the units of force. All vectors can be decomposed this way, regardless of the units associated with them. There are a lot of vectors in physics (and other subjects), and it is always useful to know how to break them up into components.
Best Answer
The above two cases are equivalent, we know that due to the couple, the disc would be at equilibrium. The torque about the center, would also be same because it is given by $\tau = F \times R = FRsin(\theta)$ and if $R$ has changed, so has the $sin(\theta)$, you will take components of the force perpendicular to the radius and the torque would remain as $R$ will become $\sqrt{2}$ and $sin(\theta)$ will become $\frac{1}{\sqrt{2}}$