The hyperbola – Conics (Senior 6 Mathematics) | Ganzaa

Conics

Lessons

Senior 6 · Mathematics

Conics

5 lessons

Lesson 4 of 5

The hyperbola

40 min

Locating a sound

Two microphones 2 km apart hear the same thunderclap, one of them 3 seconds before the other. Sound travels about 340 metres per second, so the source is 1020 m nearer to the first microphone than to the second. The points with this property do not form a single point but a curve: a branch of a hyperbola. A third microphone gives a second hyperbola, and the source is where they cross. Radio navigation systems for ships worked on the same idea for decades. This lesson studies the curve defined by a constant difference of distances.

Definition

Hyperbola: given two points $F_{1}$ and $F_{2}$ , the foci, with $F_{1} F_{2} = 2 c$ , and a length $2 a$ with $0 \lt 2a \lt 2c$ , the hyperbola is the set of points $P$ with

∣ P F_{1} - P F_{2} ∣ = 2 a .

It has two branches: $P F_{1} - P F_{2} = 2 a$ on the branch near $F_{2}$ and $P F_{2} - P F_{1} = 2 a$ on the branch near $F_{1}$ . The midpoint of $F_{1} F_{2}$ is the centre; the points where the line $F_{1} F_{2}$ meets the curve are the vertices, and the segment between them is the transverse axis (length $2 a$ ).

Important

Standard equation. With $F_{1} (- c, 0)$ , $F_{2} (c, 0)$ , the same two squarings as for the ellipse turn $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm 2a$ into $(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2})$ . Now $c \gt a$ , so put $b^{2} = c^{2} - a^{2}$ :

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

Asymptotes. Solving for $y$ : $y = \pm\frac{b}{a}\sqrt{x^2 - a^2}$ . For large $∣ x ∣$ , $\sqrt{x^2 - a^2}$ is close to $∣ x ∣$ ; precisely, $\frac{b}{a}\big(x - \sqrt{x^2 - a^2}\big) = \frac{ab}{x + \sqrt{x^2 - a^2}} \to 0$ as $x \to +\infty$ . So the curve approaches the lines

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

which are the diagonals of the rectangle $|x| \le a$ , $|y| \le b$ . The segment from $(0, - b)$ to $(0, b)$ is the conjugate axis.

Element	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$	$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
transverse axis	on the $x$ -axis	on the $y$ -axis
vertices	$(\pm a, 0)$	$(0, \pm a)$
foci	$(\pm c, 0)$ , $c = \sqrt{a^2 + b^2}$	$(0, \pm c)$
eccentricity	$e = \frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}} \gt 1$
directrices	$x = \pm\frac{a}{e}$	$y = \pm\frac{a}{e}$
asymptotes	$y = \pm\frac{b}{a}x$	$y = \pm\frac{a}{b}x$
latus rectum	length $\frac{2b^2}{a}$
parametric form	$(a\sec\theta,\ b\tan\theta)$	$(b\tan\theta,\ a\sec\theta)$

Here $a$ is under the positive term, whichever of $a$ and $b$ is larger.

The hyperbola

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

, drawn from its equation. The asymptotes are the diagonals of the dashed

2a \times 2b

rectangle;

P = \left(\frac{20}{3}, 4\right)

Worked example 1: the elements of a hyperbola

Find the vertices, foci, eccentricity, directrices and asymptotes of $9 x^{2} - 16 y^{2} = 144$ .

Solution. Divide by 144: $\frac{x^2}{16} - \frac{y^2}{9} = 1$ , so $a = 4$ , $b = 3$ and $c = \sqrt{16 + 9} = 5$ .

vertices $(\pm 4, 0)$ , foci $(\pm 5, 0)$ , $e = \frac{5}{4}$ ;
directrices $x = \pm\frac{a}{e} = \pm\frac{16}{5}$ ; asymptotes $y = \pm\frac{3}{4}x$ .

Check with the point $P\left(\frac{20}{3}, 4\right)$ of the figure: $\frac{400}{9 \times 16} - \frac{16}{9} = \frac{25}{9} - \frac{16}{9} = 1$ , so $P$ is on the curve, and $PF_1 = \sqrt{\left(\frac{35}{3}\right)^2 + 16} = \frac{37}{3}$ , $PF_2 = \sqrt{\left(\frac{5}{3}\right)^2 + 16} = \frac{13}{3}$ , whose difference is $8 = 2 a$ .

Worked example 2: a hyperbola with a vertical transverse axis

Find the hyperbola with vertices $(0, \pm 3)$ and foci $(0, \pm 5)$ , and its asymptotes.

Solution. $a = 3$ , $c = 5$ , so $b^{2} = 25 - 9 = 16$ . The transverse axis is vertical: $\frac{y^2}{9} - \frac{x^2}{16} = 1$ . Asymptotes: $y = \pm\frac{a}{b}x = \pm\frac{3}{4}x$ , the same lines as in worked example 1. The two hyperbolas are called conjugate: they share their asymptotes.

Tangent and normal; the rectangular hyperbola

Implicit differentiation of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ gives $\frac{dy}{dx} = \frac{b^2x}{a^2y}$ , and, as for the ellipse,

\text{tangent at } (x_1, y_1):\ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 , \qquad \text{normal } (x_1y_1 \ne 0):\ \frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2 .

The line $y = m x + k$ is a tangent if and only if $k^{2} = a^{2} m^{2} - b^{2}$ . This needs $|m| \gt \frac{b}{a}$ : no tangent is parallel to an asymptote, or less steep than one.

A hyperbola with $a = b$ is rectangular (or equilateral): its asymptotes $y = \pm x$ are perpendicular and $e = \sqrt{2}$ . Turned through $45^\circ$ so that the asymptotes become the coordinate axes, it has equation $x y = c^{2}$ , with parametric form $\left(ct, \frac{c}{t}\right)$ , $t \ne 0$ , and tangent at parameter $t$ : $x + t^{2} y = 2 c t$ .

Worked example 3: tangent and normal at a point

Find the tangent and the normal to $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at $P\left(\frac{20}{3}, 4\right)$ .

Solution. Tangent: $\frac{20x}{3 \times 16} - \frac{4y}{9} = 1$ , that is $\frac{5x}{12} - \frac{4y}{9} = 1$ ; multiplying by 36: $15 x - 16 y = 36$ .

Normal: $\frac{16x}{\frac{20}{3}} + \frac{9y}{4} = 25$ , that is $\frac{12x}{5} + \frac{9y}{4} = 25$ ; multiplying by 20: $48 x + 45 y = 500$ . Check: $48 \times \frac{20}{3} + 45 \times 4 = 320 + 180 = 500$ , and the slopes $\frac{15}{16}$ and $-\frac{48}{45} = -\frac{16}{15}$ multiply to $- 1$ .

Worked example 4: a rectangular hyperbola

Find the tangent to $x y = 16$ at $(8, 2)$ and the area of the triangle it forms with the coordinate axes.

Solution. $c^{2} = 16$ , $c = 4$ , and $(8, 2) = \left(4t, \frac{4}{t}\right)$ with $t = 2$ . Tangent: $x + 4 y = 16$ . Check by differentiation: $y = \frac{16}{x}$ , $y' = -\frac{16}{x^2} = -\frac{1}{4}$ at $x = 8$ , the slope of $x + 4 y = 16$ .

The tangent meets the axes at $(16, 0)$ and $(0, 4)$ , so the area is $\frac{1}{2} \times 16 \times 4 = 32 = 2c^2$ . In general the tangent at $t$ meets the axes at $(2 c t, 0)$ and $\left(0, \frac{2c}{t}\right)$ , so the area is always $2 c^{2}$ , whatever the point of contact.

Worked example 5: where did the sound come from?

Microphones $A$ and $B$ are at $(- 1, 0)$ and $(1, 0)$ , in kilometres. A sound reaches $A$ exactly 3 s before $B$ . Taking the speed of sound as $340\,\mathrm{m\,s^{-1}}$ , find the curve on which the source lies.

Solution. $PB - PA = 3 \times 0.340 = 1.02\,\mathrm{km} = 2a$ , so $a = 0.51$ . The foci are $A$ and $B$ , so $c = 1$ and $b^{2} = 1 - {0.51}^{2} = 0.7399$ . The source lies on

\frac{x^2}{0.2601} - \frac{y^2}{0.7399} = 1 , \quad x \lt 0 ,

the branch nearer to $A$ , since the sound reached $A$ first. A third microphone would give a second hyperbola through the same source.

Warning

For a hyperbola, $c^{2} = a^{2} + b^{2}$ . Pupils who carry over $c^{2} = a^{2} - b^{2}$ from the ellipse put the foci inside the vertices, where there is no curve.

The hyperbola never meets its asymptotes in the standard position, but it gets as close as you like. An asymptote is not a tangent at a point.

Tip

To sketch $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ quickly: draw the rectangle with corners $(\pm a, \pm b)$ , extend its diagonals as the asymptotes, mark the vertices $(\pm a, 0)$ , then draw each branch from its vertex towards the asymptotes. Replacing the 1 by 0 in the equation gives $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$ , the pair of asymptotes.

Khan Academy: Foci of a hyperbola (about 15 min). Watch how the foci are found from c² = a² + b², and compare with the ellipse in lesson 3.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the vertices, foci, eccentricity, directrices and asymptotes of $\frac{x^2}{9} - \frac{y^2}{16} = 1$ .
Find the hyperbola with foci $(\pm 13, 0)$ and vertices $(\pm 5, 0)$ .
Write $4 x^{2} - 9 y^{2} - 16 x - 18 y - 29 = 0$ in standard form, and give its centre and asymptotes.
Find the tangent and the normal to $\frac{x^2}{4} - \frac{y^2}{9} = 1$ at $(4, 3\sqrt{3})$ .
Find the tangents to $\frac{x^2}{16} - \frac{y^2}{9} = 1$ with slope 1. Explain why there is none with slope $\frac{1}{2}$ .
Find the tangent to $x y = 9$ at $(9, 1)$ , and check that it forms with the axes a triangle of area $2 c^{2}$ .
Prove that, for every point of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , the product of its distances to the two asymptotes is $\frac{a^2b^2}{a^2 + b^2}$ .

Show answersHide answers

$a = 3$ , $b = 4$ , $c = 5$ : vertices $(\pm 3, 0)$ , foci $(\pm 5, 0)$ , $e = \frac{5}{3}$ , directrices $x = \pm\frac{9}{5}$ , asymptotes $y = \pm\frac{4}{3}x$ .
$b^{2} = 169 - 25 = 144$ : $\frac{x^2}{25} - \frac{y^2}{144} = 1$ .
$\frac{(x - 2)^2}{9} - \frac{(y + 1)^2}{4} = 1$ : centre $(2, - 1)$ , asymptotes $y + 1 = \pm\frac{2}{3}(x - 2)$ .
Tangent $\sqrt{3}x - y = \sqrt{3}$ ; normal $x + \sqrt{3}y = 13$ .
$k^{2} = 16 - 9 = 7$ : $y = x \pm\sqrt{7}$ . With $m = \frac{1}{2}$ , $k^2 = 4 - 9 \lt 0$ : impossible, since $\frac{1}{2} \lt \frac{b}{a} = \frac{3}{4}$ .
$c = 3$ , $t = 3$ : $x + 9 y = 18$ ; intercepts 18 and 2, area $18 = 2 \times 9$ .
The asymptotes are $b x - a y = 0$ and $b x + a y = 0$ . The product of the distances is $\frac{|b^2x^2 - a^2y^2|}{a^2 + b^2} = \frac{a^2b^2}{a^2 + b^2}$ , since $b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2}$ on the curve.

Sources and credits

Written by Ganzaa: the explanations, the derivations, the diagram (plotted from $\frac{x^2}{16} - \frac{y^2}{9} = 1$ with the computed tangent), the worked examples, the practice items and the quiz questions.
Photo: Thtr300 kuehlturm, Smial, Wikimedia Commons, CC BY-SA 2.0 de.
Video: Foci of a hyperbola, Khan Academy, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Locating a sound

2Tangent and normal; the rectangular hyperbola

Locating a sound

Tangent and normal; the rectangular hyperbola