geometric-opticsopticsreflectionvisible-light
Suppose a ray of light hits a concave mirror and is parallel to the principal axis but far away from it such that it doesn't follow paraxial ray approximation. Will it pass through focus or between focus and radius of curvature or between pole and focus?
Here pole, focus and radius of curvature mean the same thing as in paraxial ray approximation .
As Olin said, for a parabolic mirror lines that start out parallel to the principal axis all converge to a point. This can easily be confirmed with a bit of simple math. For a mirror, angle of incidence = angle of reflection. I show that this results in a parabolic shape in my answer to an earlier question (see the last part of the answer).
Now a spherical surface is a reasonable approximation for a parabolic surface when the angles are small - the error gives rise to spherical aberration which limits the performance of (especially large) spherical optical elements, but it's a small error for small lenses. You can convince yourself by writing down the difference:
$$y = a x^2\tag{parabolic}$$
$$(r - y)^2 + x^2 = r^2 \tag{spherical}$$
We can rewrite the spherical equation as
$$r^2 - 2ry + y^2 + x^2 = r^2\\ y = \frac{x^2}{2r}+\frac{y^2}{2r}$$
And when $y$ is very small, the first term dominates. However, at larger distances off axis, the second term becomes significant and the spherical surface is no longer a good approximation of the parabolic one.
As for the last part of your question - your own sketch effectively shows the answer. The tangent to the mirror surface is perpendicular to the ray from center of curvature. The laws of reflection tell you that incident =reflected angle, so for small angles the ray goes throught the point that is midway between the mirror and the center of curvature.
The ray passing from the focus, after the lens overlaps with the one that starts from the tip of the virtual image, see the image. (Basically that ray is not a "special" one).
Best Answer
Assuming mirror to be spherical section. C is the center of sphere.
See, Using trigonometry. $$x=d \times \sin(2\theta)$$ $$x=R\times\sin\theta$$
Eliminate $\theta$ and get $d$ : distance from Center of curvature as a function of $x$.
Verify for small theta where $\sin\theta\approx\theta$
If you just want to see that which side ray bents then see. $$d=\dfrac{R\times \sec\theta}2$$ which shows that $d\ge R/2$ . So, ray bends towards the pole as looses it paraxial character.