Hyperbolas, calculus lesson

Two stations send a pulse at the same moment, and a ship hears one of them a little earlier than the other. The time gap tells the ship how much closer it is to one station, but not where it is. All the places with that same difference in distance form a hyperbola, and older navigation systems found ships by crossing two such curves.

Key terms

Difference of focal distances

Definition Let $F_1$ and $F_2$ be two points with $\lvert F_1F_2 \rvert = 2c > 0$ , and let $0 < a < c$ . The hyperbola with foci $F_1$ and $F_2$ is the set of points $P$ with

\big\lvert \, \lvert PF_1 \rvert - \lvert PF_2 \rvert \, \big\rvert = 2a.

The condition $a < c$ is needed, because by the triangle inequality the difference of the two distances can never exceed $2c$ , and it equals $2c$ only on the two rays of the line $F_1F_2$ that start at the foci and point away from each other.

In plain words For every point on the curve, one focus is farther than the other by the same fixed amount $2a$ .

Example Take foci $(\pm 5, 0)$ and $2a = 6$ . The point $(3, 0)$ is $8$ from $(-5, 0)$ and $2$ from $(5, 0)$ , and $8 - 2 = 6$ , so it lies on the hyperbola.

Foci and the standard equation

Definition With foci $(\pm c, 0)$ and $0 < a < c$ , set $b = \sqrt{c^2 - a^2} > 0$ . The hyperbola is

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad c^2 = a^2 + b^2.

With foci $(0, \pm c)$ it is

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1.

The variable with the positive sign tells you which axis the foci are on. Unlike the ellipse, $a$ need not be larger than $b$ .

In plain words Same shape of equation as the ellipse but with a minus sign, and now $c^2 = a^2 + b^2$ , so the foci are farther out than the vertices.

Example For foci $(\pm 5, 0)$ and $a = 3$ , $b^2 = 25 - 9 = 16$ , so the hyperbola is

\frac{x^2}{9} - \frac{y^2}{16} = 1.

For $\frac{y^2}{4} - \frac{x^2}{5} = 1$ , $c = \sqrt{4 + 5} = 3$ and the foci are $(0, \pm 3)$ .

Two branches

Definition The hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has no points with $\lvert x \rvert < a$ , since there

\frac{x^2}{a^2} < 1 \le 1 + \frac{y^2}{b^2}.

It splits into two branches, the right branch with $x \ge a$ , where $\lvert PF_1 \rvert - \lvert PF_2 \rvert = 2a$ for $F_1 = (-c, 0)$ and $F_2 = (c, 0)$ , and the left branch with $x \le -a$ , where $\lvert PF_2 \rvert - \lvert PF_1 \rvert = 2a$ .

In plain words The curve comes in two separate pieces, each one wrapped around its own focus.

Example For $x^2 - y^2 = 1$ , solving for $y$ gives $y = \pm\sqrt{x^2 - 1}$ , which is real only for $\lvert x \rvert \ge 1$ . The vertical strip $-1 < x < 1$ contains no points of the curve.

Transverse and conjugate axes

Definition For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , the vertices are $(\pm a, 0)$ . The transverse axis is the segment between the vertices, of length $2a$ , and it lies on the line through the foci. The conjugate axis is the segment from $(0, -b)$ to $(0, b)$ , of length $2b$ . They meet at the center.

In plain words The transverse axis joins the two branches at their tips. The conjugate axis crosses it at right angles and does not meet the curve, but it sets the slopes of the asymptotes.

Example Divide $9x^2 - 4y^2 = 36$ by $36$ to get $\frac{x^2}{4} - \frac{y^2}{9} = 1$ . The vertices are $(\pm 2, 0)$ , the transverse axis has length $4$ and the conjugate axis has length $6$ .

Eccentricity

Definition The eccentricity of a hyperbola is

e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} > 1.

In plain words The larger $e$ , the wider the branches open. Close to $1$ they are narrow.

Example For $\frac{x^2}{4} - \frac{y^2}{9} = 1$ , $c = \sqrt{13}$ and $e = \frac{\sqrt{13}}{2} \approx 1.80$ .

Directrices

Definition For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $e = \frac{c}{a}$ , the directrices are the lines

x = \pm \frac{a}{e} = \pm \frac{a^2}{c}.

Every point $P$ on the hyperbola satisfies $\lvert PF \rvert = e \cdot \operatorname{dist}(P, L)$ , where $F = (c, 0)$ goes with $L$ the line $x = \frac{a}{e}$ , and $F = (-c, 0)$ goes with $x = -\frac{a}{e}$ . Since $e > 1$ , the directrices lie between the two vertices.

In plain words This is the focus directrix definition with $e > 1$ . Each point on the curve is $e$ times as far from a focus as from the matching line.

Example For $\frac{x^2}{9} - \frac{y^2}{16} = 1$ , $c = 5$ and $e = \frac{5}{3}$ , so the directrices are $x = \pm \frac{9}{5}$ . The vertex $(3, 0)$ is $2$ from $(5, 0)$ and $\frac{6}{5}$ from $x = \frac{9}{5}$ , and $\frac{5}{3} \cdot \frac{6}{5} = 2$ .

Asymptotes

Definition The hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has the asymptotes

y = \frac{b}{a}x \quad \text{and} \quad y = -\frac{b}{a}x.

On the part of the right branch with $y > 0$ we have $y = \frac{b}{a}\sqrt{x^2 - a^2}$ , and the vertical gap to the line $y = \frac{b}{a}x$ is

\frac{b}{a}x - \frac{b}{a}\sqrt{x^2 - a^2} = \frac{b}{a} \cdot \frac{a^2}{x + \sqrt{x^2 - a^2}} \to 0 \quad \text{as } x \to \infty.

The other three parts work the same way by symmetry. For $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the asymptotes are $y = \pm \frac{a}{b}x$ .

In plain words Far from the center the branches look almost like straight lines. A quick sketch starts with the rectangle with corners $(\pm a, \pm b)$ and draws its diagonals.

Example For $\frac{x^2}{9} - \frac{y^2}{16} = 1$ the asymptotes are $y = \pm \frac{4}{3}x$ . At $x = 30$ the upper right part has $y = \frac{4}{3}\sqrt{891} \approx 39.80$ , only about $0.20$ below the asymptote value $40$ .

Reflection property

Definition Let $P$ be a point on a hyperbola with foci $F_1$ and $F_2$ . The tangent line at $P$ bisects the angle $F_1PF_2$ between the segments $PF_1$ and $PF_2$ . So a ray of light traveling toward $F_2$ that meets the branch around $F_2$ on its outer, convex side reflects along the line toward $F_1$ , and likewise with the two foci exchanged.

In plain words The tangent splits the angle between the segments to the two foci exactly in half. Some telescopes pair a hyperbolic mirror with a parabolic one for this reason.

Example On $\frac{x^2}{9} - \frac{y^2}{16} = 1$ take $P = \left(5, \frac{16}{3}\right)$ . With $F_1 = (-5, 0)$ and $F_2 = (5, 0)$ the focal distances are

\lvert PF_2 \rvert = \frac{16}{3}, \qquad \lvert PF_1 \rvert = \sqrt{100 + \frac{256}{9}} = \frac{34}{3}.

The tangent $\frac{5x}{9} - \frac{y}{3} = 1$ meets the $x$ axis at $T = \left(\frac{9}{5}, 0\right)$ , and

\frac{\lvert F_1T \rvert}{\lvert TF_2 \rvert} = \frac{34/5}{16/5} = \frac{17}{8} = \frac{\lvert PF_1 \rvert}{\lvert PF_2 \rvert}.

By the angle bisector theorem, the line $PT$ bisects the angle at $P$ .

Check yourself

Find the foci and asymptotes of $\frac{x^2}{16} - \frac{y^2}{9} = 1$ .

Show the answer

Here $c = \sqrt{16 + 9} = 5$ , so the foci are $(\pm 5, 0)$ . The asymptotes are $y = \pm \frac{3}{4}x$ .

A hyperbola has vertices $(\pm 2, 0)$ and foci $(\pm 4, 0)$ . Write its equation.

Show the answer

The foci are on the $x$ axis with $a = 2$ and $c = 4$ , so $b^2 = 16 - 4 = 12$ , and the equation is

\frac{x^2}{4} - \frac{y^2}{12} = 1.

Find the eccentricity of $x^2 - y^2 = 1$ .

Show the answer

Here $a = b = 1$ , so $c = \sqrt{2}$ and $e = \sqrt{2}$ .

Hyperbolas

Key terms

Difference of focal distances

Foci and the standard equation

Two branches

Transverse and conjugate axes

Eccentricity

Directrices

Asymptotes

Reflection property

Check yourself

How this lesson connects

Builds on

Used later in