The hyperbola: a constant difference of distances, and its asymptotes

Hyperbola

What stays constant on a hyperbola, and why do its branches hug the lines y = ±(b/a)x without touching them?

Drag a point along the right branch: |PF₁| − |PF₂| stays at 2a while the gap to the asymptote shrinks toward zero. The dashed rectangle with corners (±a, ±b) gives the asymptotes as its diagonals and c as its half-diagonal.

Drag the ringed points on the board.

High school
Definition by two foci
Standard equation
c² = a² + b²
Asymptotes
Eccentricity e = c/a

a

2.0

b

1.5

y_P

3.0

\big||PF_1| - |PF_2|\big| = 2a

$|PF_1|$: 7.590
$|PF_2|$: 3.590
$|PF_1| - |PF_2|$: 4.000
$c = \sqrt{a^2 + b^2}$: 2.500
$e = c / a$: 1.250
$\pm\dfrac{b}{a}$: 0.750
$\text{gap to asymptote}$: 0.354

Guided tour

1The difference is 2a
With $a=2$ , $b=1.5$ the foci are at $(\pm2.5,0)$ . At $P$ with $y_P=1.5$ : $|PF_1|=5.536$ and $|PF_2|=1.536$ , so $|PF_1|-|PF_2|=4=2a$ . Drag $P$ along the branch and the difference holds. On the left branch it is $|PF_2|-|PF_1|$ that equals $2a$ .
2Closer and closer to the asymptote
The red segment is the vertical gap between $P$ and the line $y=\frac{b}{a}x$ . At $y_P=1.5$ it is $0.621$ ; drag $P$ out to $y_P=5$ and it shrinks to $0.220$ . It keeps shrinking toward $0$ but never reaches it.
3The rectangle holds a, b and c
Put $P$ at the vertex $(2,0)$ : $|PF_1|=4.5$ , $|PF_2|=0.5$ . The dashed rectangle has corners $(\pm2,\pm1.5)$ ; its diagonals are the asymptotes, slope $\pm0.75$ , and its half-diagonal is $c=\sqrt{2^2+1.5^2}=2.5$ . The circle of radius $c$ through the corners passes through both foci. So $e=\frac{c}{a}=1.25$ .
4a = b: a rectangular hyperbola
Set $b=2$ as well. The rectangle becomes a square, the asymptotes $y=\pm x$ are perpendicular, and $e=\sqrt2\approx1.414$ . Raise $b$ further and the branches open wider; $e$ grows with them.

Explanation

1
Definition and equation
A hyperbola is the set of points whose distances to two foci $F_1$ , $F_2$ differ by a constant: $\big||PF_1|-|PF_2|\big|=2a$ , with $0<2a<|F_1F_2|=2c$ . (By the triangle inequality the difference can never exceed $2c$ .)
With $F_1=(-c,0)$ , $F_2=(c,0)$ , the same two squarings as for the ellipse give $(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$ . Now $c>a$ , so name $b^2=c^2-a^2$ :
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\qquad(a>0,\ b>0)$
2
Ellipse and hyperbola side by side
- Ellipse: sum $2a$ , so $2a>2c$ and $a^2=b^2+c^2$ ; $a$ is the longest of the three.
- Hyperbola: difference $2a$ , so $2a<2c$ and $c^2=a^2+b^2$ ; $c$ is the longest.
One sign in the definition flips the sign in the equation and decides which length is the hypotenuse.
3
Why the asymptotes are y = ±(b/a)x
Solve for the upper half of the right branch: $y=\frac{b}{a}\sqrt{x^2-a^2}$ . Its vertical gap below the line $y=\frac bax$ is
$\frac ba\left(x-\sqrt{x^2-a^2}\right)=\frac ba\cdot\frac{a^2}{x+\sqrt{x^2-a^2}}=\frac{ab}{x+\sqrt{x^2-a^2}}$
The gap is always positive, so the curve never touches the line, and it tends to $0$ as $x$ grows. A quick way to recall them: replace the $1$ in the equation by $0$ , $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ , which factors into the two lines.
4
Eccentricity and the opening
$e=\dfrac ca>1$ , and $\dfrac ba=\sqrt{e^2-1}$ . The larger $e$ , the steeper the asymptotes and the wider the branches open. At $e=\sqrt2$ , $a=b$ and the asymptotes are perpendicular: the rectangular hyperbola, which turned through $45^\circ$ is the familiar $y=\frac kx$ .

Check yourself

Q01

What are the asymptotes of $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ ?

Q02

For $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ , what are $c$ and $e$ ?

Q03

A point $P$ on the right branch of $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ has $|PF_1|=10$ . What is $|PF_2|$ ?

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