Hall & Higson define an ellipse as:
Pick any two points. These will be called the foci of the ellipse. The set of all points at which the sum of the distances to the two foci is some fixed number is an ellipse. Equivalently, affix two tacks to a board, tie each end of a string to a tack, and draw the curve created by a pencil which stretches the string taut.
Exercise 3(a) is to show that the foci of the ellipse must lie on the semimajor axes. Why is this?
Note that the endpoints of the semiminor axis $A$ and $B$ are at equal distance from each foci $f_1$ and $f_2$, so the distance from $f_1$ to $A$ is the same that de distance from $f_1$ to B, therefore $f_1$ must lie on the perpendicular bisector of $A$ and $B$, which is the semimajor axis. The same argument works for $f_2$.
Edit: I think this is an easier argument: from the tack-and-string definition we get a triangle with fixed base $\overline{f_{1}f_{2}}$ and sides $a$, $b$ such that $a+b=\text{constant}$. The endpoints of the semiminor axis are the points where the height of this triangle is maximum, and the endpoints of the semimajor axis are the points where the height of this triangle is minimum, in this case the minimum is 0. If the height of the triangle is zero this means that the top vertex is collineal with the vertex of the base, and this implies that the foci are in the semimajor axis
Let $F_1$, $F_2$ be the foci of the ellipse, $O$ its center (midpoint of $F_1F_2$). Any point $P$ of the ellipse has the sum of its distances from the foci which is a fixed number $2a$ (greater than $F_1F_2$).
If $P$ doesn't lie on line $F_1F_2$, then in triangle $PF_1F_2$ the median $PO$ is less then ${1\over2}F_1F_2$ (see here for a proof), that is $PO<a$.
On the other hand, if $P$ lies on the line of the foci and we call $F_1$ the focus nearest to $P$, we have: $$2a=PF_1+PF_2=2PF_1+F_1F_2=2PF_1+2F_1O=2PO,$$ that is $PO=a$. Of course there are two such points, symmetric about $O$.
Hence the points on the ellipse farthest from the center lie on the line of the foci, and the segment between them is the longest diameter of the ellipse, whence the name major axis.
Let the sum of two distances from the tacks to and through tracing point be $2a$, and let their difference be $2c$.
$a>c$ always.
Call the line drawn between the two tacks as major line. This is a symmetry axis of the figure, the ellipse.
Similarly draw a perp bisector to the inter-tacks line as a minor line. This is also another second symmetry axis. Call this symmetry axis as minor line.
Draw the Pythagorean central right triangle of sides $( a,c,\sqrt{a^2-c^2}).$ The central "diameters " are $( 2a,2\sqrt{a^2-c^2}).$
It an accepted practice to call the bigger portion/segment of such inter-tack line contained between extremities as the major axis and the smaller one as the minor axis.
Further canonical form of an ellipse representation is:
$$\frac{x^2}{a^2}+\frac{ y^2}{(a^2-c^2)} =1;$$
so actually, if an ellipse minor axis length is greater than the major axis, then the half inter-focal segment length $c$ has to be a pure imaginary number, i.e., $c^2<0, $ a matter that had been conveniently and continuously overlooked.
