# How to prove the foci of the ellipse must lie on the semimajor axes if the tack-and-string definition is true

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?

• Ellipse, not eclipse. Ellipse is an oval shape. Eclipse is when one heavenly body crosses between two others leaving a shadow on one. Commented Jun 9, 2021 at 12:37
• "if the tack-and-string definition is true" hasn't any meaning in mathematics; A definition is neither true nor false. You should have said: "Let us take the tack-and-string definition of the ellipse."... Commented Jun 9, 2021 at 20:30
• This is obvious from the symmetry, isn't it? The figure drawn by the tack-and-string method is clearly symmetrical about the line joining the tacks. Commented Jun 11, 2021 at 9:20
• What is your definition of "the semimajor axes"? Commented Jul 5 at 13:31

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

• I'm not sure I follow the argument. Why is the endpoints of the semiminor axis A and B are at equal distance from each foci f1 and f2?This doesn't seem to follow from the tack-and-string definition. It is also very clear taht semiminor and semimajor axis are perpendicular bisector to each other. I mean, yeah that is true but cannot be direcly followed from the definition. Commented Jun 9, 2021 at 14:06
• 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 $a+b=$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.
– Nah
Commented Jun 9, 2021 at 15:22
• "the endpoints of the semimajor axis are the points where the height of this triangle is minimum" I don't see that as obvious? Commented Jun 11, 2021 at 10:32

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.

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.

• It is not a convention: major axis is longer than minor axis. Commented Jul 5 at 13:18
• Yes, it is what is meant, edited. Commented Jul 5 at 17:08
• But then you cannot write "if an ellipse minor axis length is greater than the major axis". Commented Jul 5 at 17:39