How a Fixed Difference of Distances Creates a Curve With Two Branches

An ellipse keeps the sum of the distances to two foci fixed; a hyperbola keeps their difference fixed. That small change produces a curve with two separate branches that stretch away without end, approaching a pair of straight lines.

This part covers the standard equation and eccentricity, the features of a hyperbola, and its focal property.

**A hyperbola with centre at the origin and transverse axis along the x-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with $b^2=a^2(e^2-1)$ and $e>1$; when the $x^2$ term is the negative one, as in $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$, the transverse axis is vertical.

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

Two cases:

- $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ — opens left and right, $e=\sqrt{1+\frac{b^2}{a^2}}$
- $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ — opens up and down, $e=\sqrt{1+\frac{a^2}{b^2}}$

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

$$e=\sqrt{1+\frac{9}{16}}=\frac{5}{4}=1.25$$

Worked example 2. For $9y^2-16x^2=144$, dividing by 144 gives $\frac{y^2}{16}-\frac{x^2}{9}=1$. The transverse axis is vertical, and $e=\sqrt{1+\frac{9}{16}}=1.25$.

An everyday example. A table lamp with an open cylindrical shade throws light on a nearby wall with a bright edge shaped like a hyperbola.

The substance. **Unlike an ellipse, a hyperbola may have $b>a$** — the sign of each term, not its size, decides which axis is transverse.

**For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the centre is the origin, the vertices are $(\pm a,0)$, the foci are $(\pm ae,0)$, the transverse and conjugate 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 $\frac{x^2}{16}-\frac{y^2}{9}=1$, with $a=4$, $b=3$ and $e=1.25$:

- Vertices $(\pm 4,0)$
- Foci $(\pm 5,0)$, since $ae=5$
- Transverse axis 8 and conjugate axis 6
- Latus rectum $\frac{2\times 9}{4}=4.5$
- Directrices $x=\pm \frac{4}{1.25}=\pm 3.2$
- Asymptotes $y=\pm \frac{3}{4}x$

Sketching. Draw the rectangle with sides $2a$ and $2b$ centred at the origin, extend its diagonals as the asymptotes, and draw the two branches through the vertices, bending towards the asymptotes.

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

An everyday example. The cooling towers of a thermal power station have a narrow waist whose outline, seen side-on, is a hyperbola.

The substance. The directrices lie between the centre and the vertices — here at 3.2, inside the vertices at 4, the opposite of an ellipse.

**For every point on a hyperbola, the difference of its distances from the two foci is constant and equal to the length of the transverse axis: $|P{S}^{\prime }-PS|=2a$.

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

Worked example. For $\frac{x^2}{16}-\frac{y^2}{9}=1$ and the point $P(5,2.25)$ on it:

- $PS=1.25\times 5-4=2.25$ and $P{S}^{\prime }=1.25\times 5+4=10.25$
- $P{S}^{\prime }-PS=8=2a$

Direct check: the distance from $(5,2.25)$ to $(5,0)$ is 2.25, and to $(-5,0)$ it is $\sqrt{100+5.0625}=10.25$.

Worked example 2. A point moves so that the difference of its distances from $(\pm 5,0)$ is 6. Then $2a=6$, $ae=5$ and $b^2=25-9=16$, so its path is $\frac{x^2}{9}-\frac{y^2}{16}=1$.

An everyday example. Two listening posts that hear the same thunderclap at slightly different moments know the difference in their distances from the lightning, so the strike lies on one branch of a hyperbola with the posts as foci.

The substance. The constant difference must be less than the distance between the foci — otherwise no point satisfies the condition.