Suppose a disk in 3D space. Suppose that you are at the origin, and the line connecting you and the center of the ellipse generates an angle of $\theta$ with the normal to the disk. It is clear that the disk will appear as an ellipse. I have a physics textbook that states that the ratio between the axes of the apparent ellipse is $$ \frac{b}{a} = \cos\theta $$
How is this derived?
That relation is only approximate. Diagram below represents a section $AB$ of a disk of radius $a$ (blue) along the plane formed by the line connecting $O$ with the center $M$ of the disk and the normal $ME$ to the disk. The orange segment $CD$ (perpendicular to $OM$) is the ellipse minor axis: if $O$ is very far from $M$ (that is $OM\gg a$) then the dotted rays can be considered approximately parallel between them (and with $OM$), the angles formed by $CD$ with the dotted rays are approximately right and each half-segment measures $b\approx a\cos\theta$.
To get the exact relation, notice that $CM:AH=OM:OH$, thus: $$ b_1=a\cos\theta{d\over d-a\sin\theta} \quad\hbox{and in the same way:}\quad b_2=a\cos\theta{d\over d+a\sin\theta}, $$ where we set $d=OM$. As $b_1+b_2=2b$, by adding the above equalities we obtain: $$ b=a\cos\theta{d^2\over d^2-a^2\sin^2\theta}. $$
EDIT.
It is worth mentioning that even the ellipse semimajor axis is not exactly $a$, because $a$ is the ellipse width at point $M$, which is not the midpoint of $CD$. The distance between $M$ and $CD$ midpoint is in fact $$ b_1-b={a^2d\sin\theta\cos\theta\over d^2-a^2\sin^2\theta}. $$ If $a'$ is the ellipse semimajor axis we must have then $a^2/a'^2+(b_1-b)^2/b^2=1$, whence: $$ a'={a\over\sqrt{1-(a/d)^2\sin^2\theta}}. $$
EDIT 2.
Let me point out that the apparent shape is always exactly an ellipse. That happens because the apparent curve is the intersection between the circular cone projected by $O$ and a plane. The center of the ellipse, as I noticed above, is not however the projection of the disk center.