Why Every Point on an Ellipse Has the Same Total Distance to Two Pins

Planets, comets and many satellites move in ellipses, and whispering galleries use the reflecting property of an elliptical surface. An ellipse looks like a stretched circle, but its two foci give it properties a circle lacks.

This part covers the standard equation and eccentricity, the features of an ellipse, equations from given data, and the focal property.

**An ellipse with centre at the origin and major axis along the x-axis is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, and its semi-axes and eccentricity are linked by $b^2=a^2(1-e^2)$; when $b>a$ the major axis is vertical and the roles of a and b swap.

Derivation idea.** With focus $(ae,0)$ and directrix $x=\frac{a}{e}$, the condition $PS=e\times PM$ simplifies to $\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$.

Two cases:

- $a>b$ — major axis horizontal, $e=\sqrt{1-\frac{b^2}{a^2}}$
- $b>a$ — major axis vertical, $e=\sqrt{1-\frac{a^2}{b^2}}$

Worked example. For $\frac{x^2}{25}+\frac{y^2}{16}=1$: $a=5$ and $b=4$, so

$$e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}=0.6$$

Worked example 2. For $\frac{x^2}{9}+\frac{y^2}{25}=1$ the major axis is vertical and $e=\sqrt{1-\frac{9}{25}}=0.8$.

An everyday example. The water surface in a round glass tumbler tilted to one side is an ellipse — the more the glass tilts, the larger the eccentricity.

The substance. Eccentricity measures flatness — $e=0$ gives a circle, and e close to 1 gives a long, thin ellipse.

**For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, the centre is the origin, the vertices are $(\pm a,0)$, the foci are $(\pm ae,0)$, the major and minor axes have lengths $2a$ and $2b$, the latus rectum has length $\frac{2b^2}{a}$, and the directrices are $x=\pm \frac{a}{e}$.

Worked example.** For $\frac{x^2}{25}+\frac{y^2}{16}=1$, with $a=5$, $b=4$ and $e=0.6$:

- Vertices $(\pm 5,0)$; ends of the minor axis $(0,\pm 4)$
- Foci $(\pm 3,0)$, since $ae=3$
- Major axis 10 and minor axis 8
- Latus rectum $\frac{2\times 16}{5}=6.4$
- Directrices $x=\pm \frac{5}{0.6}=\pm \frac{25}{3}$

Sketching. Mark the four ends of the axes, draw a smooth oval through them, and mark the foci on the major axis inside the curve.

Useful check. $a^2=b^2+(ae{)}^2$, and here $25=16+9$.

An everyday example. An elliptical athletics track in a stadium has its two foci on the long axis, and the curve is flattest at the ends of the short axis.

The substance. For a vertical ellipse every feature moves to the y-axis — for $\frac{x^2}{9}+\frac{y^2}{25}=1$ the foci are $(0,\pm 4)$.

**Given a focus, a directrix and the eccentricity, write $PS=e\times PM$ for a point $P(x,y)$ and simplify; given data in standard position, such as foci and vertices, find a, b and e from $b^2=a^2(1-e^2)$ and substitute into the standard form.

Worked example (focus and directrix).** Find the ellipse with focus $(1,0)$, directrix $x=4$ and $e=\frac{1}{2}$.

$$\sqrt{(x-1{)}^2+y^2}=\frac{1}{2}|x-4|$$

Squaring gives $4(x^2-2x+1+y^2)=x^2-8x+16$, so $3x^2+4y^2=12$, that is

$$\frac{x^2}{4}+\frac{y^2}{3}=1$$

Check: $a=2$ and $e=\frac{1}{2}$ give a focus at $ae=1$ and a directrix at $\frac{a}{e}=4$.

Worked example 2. An ellipse has foci $(\pm 4,0)$ and vertices $(\pm 5,0)$. Then $a=5$, $ae=4$ and $b^2=25-16=9$, so the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.

An everyday example. Laying out an elliptical flower bed 10 m long with its foci 8 m apart gives exactly the second ellipse.

The substance. **Check that $e<1$** — a result with $e\ge 1$ means the data describe a parabola or a hyperbola, not an ellipse.

**For every point on an ellipse, the sum of its distances from the two foci is constant and equal to the length of the major axis, $2a$.

Why it holds.** For P on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the focal distances are $PS=a-ex$ and $P{S}^{\prime }=a+ex$, so $PS+P{S}^{\prime }=2a$.

Worked example. For $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the point $P(3,3.2)$ on it:

- $PS=5-0.6\times 3=3.2$ and $P{S}^{\prime }=5+0.6\times 3=6.8$
- $PS+P{S}^{\prime }=10=2a$

Direct check: the distance from $(3,3.2)$ to $(3,0)$ is 3.2, and to $(-3,0)$ it is $\sqrt{36+10.24}=6.8$.

Worked example 2. A point moves so that the sum of its distances from $(\pm 3,0)$ is 10. Then $2a=10$, $ae=3$ and $b^2=25-9=16$, so its path is $\frac{x^2}{25}+\frac{y^2}{16}=1$.

An everyday example. A gardener drawing an oval bed with a loop of rope around two pegs keeps the total distance to the pegs fixed, so the traced curve is an ellipse with the pegs as foci.

The substance. The constant sum must exceed the distance between the foci — if it equals that distance, the path collapses to the segment joining them.