Why a Circle Is Simply an Ellipse That Has Stopped Stretching

Both curves are built from two fixed points called foci:

- Ellipse — the set of points whose distances from the two foci add up to a constant $2a$
- Hyperbola — the set of points whose distances from the two foci differ by a constant $2a$

One rule adds, the other subtracts, and that single change explains every difference between their equations.

This part covers the equation of each curve, the relation between $a$, $b$ and $c$, eccentricity, and the key features of each.

**With foci at $(\pm c,0)$ and $P{F}_1+P{F}_2=2a$, simplifying gives $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $b^2=a^2-c^2$, so $c^2=a^2-b^2$, and the eccentricity is $e=\frac{c}{a}$ with $0\le e<1$.

Outline of the derivation.**

$$\sqrt{(x+c{)}^2+y^2}+\sqrt{(x-c{)}^2+y^2}=2a$$

Moving one root across, squaring twice and simplifying leads to

$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$$

and writing $b^2=a^2-c^2$ gives the standard form.

Worked example. An ellipse has $a=5$ and $b=3$.

$$c^2=25-9=16c=4e=\frac{4}{5}=0.8$$

An everyday example. Tie a loop of thread around two pegs on the ground and pull it tight with a stick of chalk — moving the chalk round traces an ellipse for a rangoli, because the total distance to the pegs stays fixed.

The substance. **As $e\to 0$ the foci merge and the ellipse becomes a circle**; as $e\to 1$ it flattens into a long thin shape.

**The larger denominator shows the major axis; the vertices are $a$ units from the centre along it, the foci are $c$ units along it, and the latus rectum has length $\frac{2b^2}{a}$.

Worked example 1.** $\frac{x^2}{25}+\frac{y^2}{9}=1$, with $a=5$ along the x-axis and $b=3$.

- Major axis length $10$; minor axis length $6$
- Vertices $(\pm 5,0)$; foci $(\pm 4,0)$ since $c=4$
- Eccentricity $e=\frac{4}{5}$
- Latus rectum $\frac{2\times 9}{5}=\frac{18}{5}$

Worked example 2. $\frac{x^2}{16}+\frac{y^2}{25}=1$ has its major axis along the y-axis, with $a=5$ and $b=4$.

$$c^2=25-16=9\text{vertices }(0,\pm 5)\text{foci }(0,\pm 3)e=\frac{3}{5}\text{latus rectum }\frac{32}{5}$$

Worked example 3 — from conditions. Foci $(\pm 3,0)$ and vertices $(\pm 5,0)$ give $b^2=25-9=16$:

$$\frac{x^2}{25}+\frac{y^2}{16}=1$$

To sketch, mark the centre, the four ends of the axes and the foci, then draw a smooth oval through the axis ends.

An everyday example. An oval stadium or running ground is often laid out as an ellipse, longer along one axis than the other.

The trap. **Always take $a$ from the larger denominator**, whichever variable it sits under.

**With foci at $(\pm c,0)$ and $|P{F}_1-P{F}_2|=2a$, the same algebra gives $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $b^2=c^2-a^2$, so $c^2=a^2+b^2$; since $c>a$, the eccentricity $e=\frac{c}{a}$ is greater than $1$.

Outline of the derivation.**

$$\sqrt{(x+c{)}^2+y^2}-\sqrt{(x-c{)}^2+y^2}=\pm 2a\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1$$

**Why $c>a$.** In the triangle formed by a point and the two foci, the difference of two sides is less than the third side, so $2a<2c$.

Worked example. A hyperbola has $a=3$ and $b=4$.

$$c^2=9+16=25c=5e=\frac{5}{3}$$

An everyday example. The tall cooling towers of power stations have curved walls whose outline is a hyperbola, which gives strength with less material.

The substance. **In a hyperbola, $b$ may be larger than $a$**, unlike the major and minor axes of an ellipse — the relation is $c^2=a^2+b^2$, not $a^2-b^2$.

**The variable with the positive term shows the transverse axis; the vertices are $a$ units along it, the foci are $c$ units along it, the conjugate axis has length $2b$, and the latus rectum has length $\frac{2b^2}{a}$.

Worked example 1.** $\frac{x^2}{9}-\frac{y^2}{16}=1$, with $a=3$ and $b=4$.

- Transverse axis along the x-axis, length $6$; conjugate axis length $8$
- Vertices $(\pm 3,0)$; foci $(\pm 5,0)$
- Eccentricity $e=\frac{5}{3}$
- Latus rectum $\frac{2\times 16}{3}=\frac{32}{3}$

Worked example 2. $\frac{y^2}{9}-\frac{x^2}{27}=1$ has its transverse axis along the y-axis, with $a=3$ and $b^2=27$.

$$c^2=9+27=36\text{vertices }(0,\pm 3)\text{foci }(0,\pm 6)e=2\text{latus rectum }\frac{54}{3}=18$$

Worked example 3 — from conditions. Foci $(0,\pm 13)$ and vertices $(0,\pm 5)$ give $b^2=169-25=144$:

$$\frac{y^2}{25}-\frac{x^2}{144}=1$$

An everyday example. Two radio stations sending signals at the same moment fix a ship on a hyperbola, because the difference in arrival times fixes the difference in distances.

The substance. **The two branches never meet the lines $y=\pm \frac{b}{a}x$**, which the curve approaches far from the centre.