Precalculus · G.GPE.A.3

The hyperbola

The ellipse keeps the sum of two distances fixed. What curve keeps their difference fixed instead, and why does it run off along two straight lines?

You need

The ellipsePrecalculus

Unlocks

Identifying and translating conicsPrecalculus

Where you are 6 in Conic sections

Geometry11

Conic sections6

Conic sections from a double cone Focus and directrix The ellipse

The hyperbola

The parabola as a locus Identifying and translating conics

Why this exists

Cutting a double cone with a plane steeper than its sides gives a curve in two pieces, the hyperbola. As with the ellipse, a slice is a picture, not something to compute with. The ellipse became an equation once it was described by distances to two fixed points: their sum stays the same. Keep the two points and fix the difference instead, and the result is the hyperbola. That description gives its equation, its two straight guide lines, and a way to locate a sound from the moment it arrives at two places.

The lesson runs over two days. Day 1 is the definition, the equation centered at the origin, the parts, and the asymptotes. Day 2 moves the center to (h, k), graphs with the central box, writes equations from features, completes the square, and locates a sound from a time difference.

Definition

Fix two points F_1 and F_2, the foci, and a positive length 2a smaller than the distance F_1F_2. A hyperbola is the set of all points P in the plane with

|PF_1 - PF_2| = 2a.

The parts that go with it:

The center is the midpoint of F_1F_2. Write c for the distance from the center to each focus, so F_1F_2 = 2c and the condition says 0 < a < c.
The curve has two separate pieces, the branches: the points with PF_1 - PF_2 = 2a (nearer F_2) and those with PF_2 - PF_1 = 2a (nearer F_1).
The line through the foci meets the hyperbola at two points, the vertices. The segment joining them is the transverse axis, of length 2a.
The conjugate axis is the segment through the center, perpendicular to the transverse axis, of length 2b, where b^2 = c^2 - a^2.

Figure 1For every point P on the hyperbola, d_1 and d_2 differ by the same number, 2a.

Intuition

Say a vertex on the axis sits v from the center, between the center and F_2. Its distances to the foci are c + v and c - v, which differ by 2v. That difference must be 2a, so v = a: the vertices are a from the center and the transverse axis has length 2a.

In the ellipse the sum 2a was longer than the gap 2c, so a > c. Here the difference is shorter than the gap, so c > a, and the right triangle flips: c becomes the hypotenuse. Build a rectangle centered at the center, 2a along the transverse axis and 2b across it, the central box. Its half-diagonal has length \displaystyle \sqrt{a^2 + b^2}, which is c, so a circle through the four corners passes through both foci (Figure 2):

c^2 = a^2 + b^2.

Far from the center, a point of the hyperbola is so far from both foci that the segments to them are nearly parallel, and the curve looks almost straight. Those nearly straight runs follow the box’s two diagonals, extended: the asymptotes, lines the branches approach but never touch.

Figure 2The central box has sides 2a and 2b. Its diagonals are the asymptotes, and its half-diagonal c reaches a focus, so c^2 = a^2 + b^2.

Results

Center at the origin. With the foci on the x-axis at (\pm c, 0), the hyperbola is

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

with vertices (\pm a, 0), foci (\pm c, 0) where c^2 = a^2 + b^2, and asymptotes \displaystyle y = \pm\frac{b}{a}x. With the foci on the y-axis the roles of x and y swap:

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

with vertices (0, \pm a), foci (0, \pm c) and asymptotes \displaystyle y = \pm\frac{a}{b}x. In both, a^2 sits under the positive term, whichever denominator is larger, and that variable names the direction in which the branches open. Unlike the ellipse, a may be smaller than b.

The asymptotes from the equation. Replace the 1 by 0 and solve. From \displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 comes \displaystyle y = \pm\frac{b}{a}x, and from \displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 0 comes \displaystyle y = \pm\frac{a}{b}x.

Center at (h, k). Replacing x by x - h and y by y - k slides every point by h across and k up, so

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

opens left and right, with asymptotes \displaystyle y - k = \pm\frac{b}{a}(x - h), and

\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1

opens up and down, with asymptotes \displaystyle y - k = \pm\frac{a}{b}(x - h). Vertices and foci are those of the centered hyperbola, each moved by (h, k).

Eccentricity. The ratio \displaystyle e = \frac{c}{a} is greater than 1 for every hyperbola, since c > a. The ellipses are exactly the conics with e < 1.

Why it is true

The equation. Put F_1 = (-c, 0) and F_2 = (c, 0). By the The distance formula, P = (x, y) is on the hyperbola exactly when \displaystyle \sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm 2a, that is,

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

Squaring both sides and cancelling x^2, y^2 and c^2 from each side leaves \displaystyle 4cx - 4a^2 = \pm 4a\sqrt{(x - c)^2 + y^2}. Divide by 4 and square again:

\begin{aligned} c^2x^2 - 2a^2cx + a^4 &= a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2 \\ (c^2 - a^2)x^2 - a^2y^2 &= a^2(c^2 - a^2). \end{aligned}

Since c > a, the number c^2 - a^2 is positive; call it b^2. Dividing by a^2b^2 gives \displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.

Squaring can add solutions, so check backward. On the curve, \displaystyle y^2 = (c^2 - a^2)\left(\frac{x^2}{a^2} - 1\right). Substituting into PF_2^2 = (x - c)^2 + y^2 and simplifying gives

PF_2^2 = \frac{c^2x^2}{a^2} - 2cx + a^2 = \left(\frac{cx}{a} - a\right)^2,

and in the same way \displaystyle PF_1^2 = \left(\frac{cx}{a} + a\right)^2. Every point of the curve has x^2 \ge a^2. For x \ge a both brackets are positive, so PF_1 - PF_2 = 2a; for x \le -a both are negative, and PF_2 - PF_1 = 2a. So every point of the equation satisfies the distance condition, and the right branch is the one nearer F_2.

The asymptotes. Take the upper right part of the curve, \displaystyle y = \frac{b}{a}\sqrt{x^2 - a^2} for x \ge a, and compare it with the line \displaystyle y = \frac{b}{a}x. Multiplying by the conjugate,

\begin{aligned} \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}} \\ &= \frac{ab}{x + \sqrt{x^2 - a^2}}. \end{aligned}

The gap is always positive, so the curve stays below the line. Its denominator grows without bound as x grows, so the gap shrinks toward 0. For \displaystyle \frac{x^2}{9} - \frac{y^2}{4} = 1 the gap is \displaystyle \frac{6}{x + \sqrt{x^2 - 9}}:

x	6	30	300
gap	0.536	0.100	0.010

Symmetry in both axes carries this to the other three parts of the curve.

Worked examples

Example 1

Day 1: reading the parts

Find the vertices, foci and asymptotes of \displaystyle \frac{x^2}{144} - \frac{y^2}{25} = 1, and the lengths of its axes.

Solution

The positive term holds x^2, so the branches open left and right with a^2 = 144, a = 12, and b^2 = 25, b = 5. Then c^2 = a^2 + b^2 = 169, so c = 13.

Vertices (\pm 12, 0), foci (\pm 13, 0), asymptotes \displaystyle y = \pm\frac{5}{12}x. The transverse axis has length 2a = 24 and the conjugate axis 2b = 10.

Check with the definition at x = 13: there \displaystyle y^2 = 25\left(\frac{169}{144} - 1\right) = \frac{625}{144}, so \displaystyle \left(13, \frac{25}{12}\right) is on the curve. Its distance to (13, 0) is \displaystyle \frac{25}{12}; to (-13, 0) it is \displaystyle \sqrt{26^2 + \frac{625}{144}} = \frac{313}{12}. The difference is \displaystyle \frac{288}{12} = 24 = 2a.

Example 2

Day 1: an equation to rearrange first

Find the vertices, foci and asymptotes of 4y^2 - 5x^2 = 20.

Solution

Standard form has 1 on the right, so divide both sides by 20:

\frac{y^2}{5} - \frac{x^2}{4} = 1.

The positive term holds y^2, so the branches open up and down with a^2 = 5 and b^2 = 4. Here 5 also happens to be the larger denominator, but the sign is what decides, not the size. So \displaystyle a = \sqrt{5}, b = 2, and c^2 = 5 + 4 = 9, c = 3.

Vertices \displaystyle (0, \pm\sqrt{5}), foci (0, \pm 3), asymptotes \displaystyle y = \pm\frac{\sqrt{5}}{2}x. Check at the vertex \displaystyle (0, \sqrt{5}): its distances to the foci are \displaystyle 3 - \sqrt{5} and \displaystyle 3 + \sqrt{5}, which differ by \displaystyle 2\sqrt{5} = 2a.

Example 3

Day 1: the equation from the definition

Write an equation for the set of points whose distances to (-10, 0) and (10, 0) differ by 12, and give its asymptotes.

Solution

This is a hyperbola with foci (\pm 10, 0), so its center is (0, 0), it opens left and right, and c = 10. The constant difference is 2a = 12, so a = 6. Then

b^2 = c^2 - a^2 = 100 - 36 = 64,

and

\frac{x^2}{36} - \frac{y^2}{64} = 1,

with asymptotes \displaystyle y = \pm\frac{8}{6}x = \pm\frac{4}{3}x. Check at x = 10: \displaystyle y^2 = 64 \cdot \frac{64}{36}, so \displaystyle \left(10, \frac{32}{3}\right) is on the curve. Its distances to the foci are \displaystyle \frac{32}{3} and \displaystyle \sqrt{20^2 + \frac{1024}{9}} = \frac{68}{3}, which differ by \displaystyle \frac{36}{3} = 12.

Example 4

Day 2: graphing with the central box

Graph \displaystyle \frac{(y - 1)^2}{16} - \frac{(x + 2)^2}{9} = 1, marking its center, vertices, foci and asymptotes.

Solution

Reading x + 2 as x - (-2), the center is (-2, 1). The positive term holds y, so the branches open up and down with a = 4 and b = 3, and \displaystyle c = \sqrt{16 + 9} = 5.

From the center, go 4 up and down and 3 left and right: the central box has corners (-5, 5), (1, 5), (1, -3) and (-5, -3). Its diagonals, extended, are the asymptotes

y - 1 = \pm\frac{4}{3}(x + 2).

The vertices are the middles of the top and bottom of the box, (-2, 5) and (-2, -3), and the foci are 5 from the center, at (-2, 6) and (-2, -4). Draw each branch from its vertex, bending away from the box and toward the asymptotes.

Figure 3Example 4. The branches start at the top and bottom of the box and bend out toward its extended diagonals.

Example 5

Day 2: completing the square first

Find the center, vertices, foci and asymptotes of 9x^2 - 4y^2 - 36x - 24y - 36 = 0.

Solution

Group the x terms and the y terms, move the constant, and factor out each leading coefficient, as in completing the square for a circle:

9(x^2 - 4x) - 4(y^2 + 6y) = 36.

Adding 4 inside the first parentheses adds 9 \cdot 4 = 36 to the left. Adding 9 inside the second adds -4 \cdot 9 = -36, because of the factor -4 in front. The right side gets the same:

9(x - 2)^2 - 4(y + 3)^2 = 36 + 36 - 36 = 36.

Dividing by 36:

\frac{(x - 2)^2}{4} - \frac{(y + 3)^2}{9} = 1.

The center is (2, -3); the branches open left and right with a = 2, b = 3, \displaystyle c = \sqrt{13}. Vertices (0, -3) and (4, -3); foci \displaystyle (2 + \sqrt{13}, -3) and \displaystyle (2 - \sqrt{13}, -3); asymptotes \displaystyle y + 3 = \pm\frac{3}{2}(x - 2).

Example 6

Day 2: the equation from features

A hyperbola has foci (-3, 2) and (7, 2), and its asymptotes have slopes \displaystyle \pm\frac{3}{4}. Write its equation and find its vertices.

Solution

The center is the midpoint of the foci, (2, 2), and c = 5. The foci differ only in x, so the branches open left and right, and the asymptote slopes are \displaystyle \pm\frac{b}{a}. So \displaystyle b = \frac{3}{4}a, and

a^2 + \frac{9}{16}a^2 = 25, \qquad \frac{25}{16}a^2 = 25,

so a = 4 and b = 3.

The equation is

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

with vertices (-2, 2) and (6, 2), each 4 from the center along the line y = 2.

Example 7

Day 2: locating a sound from a time difference

Jacinto and Macario run two listening stations 3400 meters apart on an east to west line, Jacinto to the west. A single loud bang reaches Macario’s station 6 seconds before it reaches Jacinto’s. Take the speed of sound that day to be 340 meters per second. (a) Write an equation for the points the bang could have come from. (b) How close to Macario’s station could it have been? (c) A third report places the bang due north of Macario’s station. Where was it?

Solution

(a) Put the center halfway between the stations, so Jacinto is at (-1700, 0) and Macario at (1700, 0), with c = 1700. In 6 seconds sound travels 6 \cdot 340 = 2040 meters, so the bang was 2040 meters farther from Jacinto than from Macario: PJ - PM = 2040 = 2a, and a = 1020. Then

\begin{aligned} b^2 &= 1700^2 - 1020^2 \\ &= 2{,}890{,}000 - 1{,}040{,}400 = 1{,}849{,}600, \end{aligned}

and the bang came from the branch of

\frac{x^2}{1{,}040{,}400} - \frac{y^2}{1{,}849{,}600} = 1

with x > 0, the branch nearer Macario.

(b) The nearest point of that branch to Macario’s station is the vertex (1020, 0), which is c - a = 680 meters away.

(c) Put x = 1700. Since \displaystyle \frac{1700^2}{1{,}040{,}400} - 1 = \frac{b^2}{a^2}, the equation gives \displaystyle y^2 = \frac{b^4}{a^2}, so \displaystyle y = \frac{b^2}{a} = \frac{1{,}849{,}600}{1020}, and \displaystyle \frac{1{,}849{,}600}{1020} = \frac{5440}{3} \approx 1813 meters north. Check: Jacinto’s distance is \displaystyle \sqrt{3400^2 + \left(\frac{5440}{3}\right)^2} = \frac{11{,}560}{3}, and \displaystyle \frac{11{,}560}{3} - \frac{5440}{3} = 2040.

Two stations give one branch, not one point. A third station, Olegario’s, paired with either of the others gives a second hyperbola, and the source lies where the two meet. Lippman and Rasmussen describe LORAN, a ship navigation system used before GPS, that located ships this way from the arrival times of radio signals sent by pairs of stations.

Figure 4Example 7. The bang reached Macario first, so it came from the branch around his station.

Boundaries

The difference must be smaller than the gap. The definition needs 0 < 2a < F_1F_2. By the Triangle inequality, |PF_1 - PF_2| \le F_1F_2, with equality only on the two rays of the line F_1F_2 that start at the foci and point away from each other. So 2a = F_1F_2 gives those two rays, and 2a > F_1F_2 gives no points at all. A difference of 0 gives the perpendicular bisector of F_1F_2, a line.
a goes with the positive term. The orientation comes from the sign, not from which denominator is larger. When a = b the asymptotes are y = \pm x, at right angles.
Axes parallel to the coordinate axes. These equations describe only hyperbolas whose axes are horizontal and vertical. The graph of \displaystyle y = \frac{1}{x} is a hyperbola turned 45^\circ, with an xy term these forms do not have.
After completing the square, look at the right side. A right side of 0, as in 9(x - 2)^2 - 4(y + 3)^2 = 0, gives two crossing lines, the asymptotes themselves. A negative right side is still a hyperbola: dividing by it flips the signs, and the branches open the other way.
The location model. Example 7 assumes sound travels in straight lines at one steady speed over flat ground. Wind, temperature and terrain change the real path and speed, so the hyperbola is a model of where the source lies, as good as those assumptions.

Where this goes next

The equations here differ from the ellipse’s by one sign, and reading those signs and squared terms to tell every conic apart is identifying a conic. The eccentricity \displaystyle e = \frac{c}{a} > 1 returns in focus and directrix, where one ratio sorts every conic, with the hyperbolas as the cases above 1. The asymptotes connect to Asymptotes in general, and the rotated hyperbola \displaystyle y = \frac{1}{x} is the starting point for graphs of rational functions.

Exercises

Day 1: reading a hyperbola

Find the vertices, foci and asymptotes.

\displaystyle \frac{x^2}{81} - \frac{y^2}{40} = 1
\displaystyle \frac{y^2}{49} - \frac{x^2}{576} = 1
\displaystyle \frac{x^2}{20} - \frac{y^2}{16} = 1
25y^2 - 4x^2 = 100
x^2 - y^2 = 9

Day 1: writing the equation

Write the equation of the hyperbola centered at the origin.

Vertices (\pm 5, 0) and foci (\pm 13, 0).
Foci (0, \pm 10) and asymptotes \displaystyle y = \pm\frac{4}{3}x.
The set of points whose distances to (0, -4) and (0, 4) differ by 2.
Vertices (\pm 4, 0) and asymptotes \displaystyle y = \pm\frac{3}{2}x. Give its foci too.
Show that \displaystyle \left(5, \frac{9}{4}\right) is on \displaystyle \frac{x^2}{16} - \frac{y^2}{9} = 1 in two ways: by substituting, and by subtracting its distances to the foci.
For \displaystyle \frac{x^2}{4} - y^2 = 1, find how far the upper right part of the curve sits below the asymptote \displaystyle y = \frac{1}{2}x at x = 10 and at x = 100.

Day 2: shifted hyperbolas

Find the center, vertices, foci and asymptotes.

\displaystyle \frac{(x - 3)^2}{9} - \frac{(y + 1)^2}{16} = 1
\displaystyle \frac{(y - 2)^2}{25} - \frac{(x - 4)^2}{144} = 1
4x^2 - y^2 + 8x + 6y - 9 = 0
16y^2 - 9x^2 + 18x + 64y - 89 = 0
Give the corners of the central box and the asymptotes of \displaystyle \frac{(x + 4)^2}{36} - \frac{(y - 1)^2}{9} = 1.

Day 2: from features to equation

Vertices (-1, 3) and (7, 3), foci (-2, 3) and (8, 3).
Center (-2, -1), a vertex at (-2, 2), and asymptotes \displaystyle y + 1 = \pm\frac{3}{2}(x + 2). Give its foci too.
Foci (1, -6) and (1, 4), and a difference of distances of 8.

Day 2: applications

Two listening stations stand 2000 meters apart. A bang reaches one of them 4 seconds before the other, and sound travels 340 meters per second. (a) With the center at the origin and the stations on the x-axis, write the equation of the hyperbola. Which branch holds the source? (b) How close to the nearer station could the source be?
Two radio stations 300 kilometers apart send a signal at the same instant. A ship receives one signal 0.0006 seconds before the other, and the signals travel 300{,}000 kilometers per second. (a) Write the equation of the hyperbola the ship is on, with the stations at (\pm 150, 0). (b) If the ship is on the line through the stations, how far is it from the nearer one?

Reasoning

Explain why the definition asks for 2a to be smaller than the distance between the foci, and describe what |PF_1 - PF_2| = 2a gives when it is not.
Show that no point of \displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 lies on either asymptote, and that the equation with 1 replaced by 0 is exactly the pair of asymptotes.
Macario says that for \displaystyle \frac{y^2}{9} - \frac{x^2}{16} = 1, a^2 = 16 because it is the larger denominator, so the vertices are (\pm 4, 0) and the foci (\pm 5, 0). Say what is right, and correct the rest.
Show that \displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 has no points with |x| < a, and explain how this splits the curve into two branches that are not bounded.

Answers

Vertices (\pm 9, 0), foci (\pm 11, 0), asymptotes \displaystyle y = \pm\frac{2\sqrt{10}}{9}x. Here a = 9, \displaystyle b = \sqrt{40} = 2\sqrt{10}, c^2 = 81 + 40 = 121.
Vertices (0, \pm 7), foci (0, \pm 25), asymptotes \displaystyle y = \pm\frac{7}{24}x. The positive term holds y^2, so a = 7, b = 24, c^2 = 49 + 576 = 625.
Vertices \displaystyle (\pm 2\sqrt{5}, 0), foci (\pm 6, 0), asymptotes \displaystyle y = \pm\frac{2\sqrt{5}}{5}x. Here c^2 = 20 + 16 = 36 and \displaystyle \frac{b}{a} = \frac{4}{2\sqrt{5}} = \frac{2\sqrt{5}}{5}.
Vertices (0, \pm 2), foci \displaystyle (0, \pm\sqrt{29}), asymptotes \displaystyle y = \pm\frac{2}{5}x. Dividing by 100 gives \displaystyle \frac{y^2}{4} - \frac{x^2}{25} = 1, so a = 2, b = 5, c^2 = 29.
Vertices (\pm 3, 0), foci \displaystyle (\pm 3\sqrt{2}, 0), asymptotes y = \pm x. Dividing by 9 gives a = b = 3, and c^2 = 18.
\displaystyle \frac{x^2}{25} - \frac{y^2}{144} = 1. Here a = 5, c = 13, b^2 = 169 - 25 = 144.
\displaystyle \frac{y^2}{64} - \frac{x^2}{36} = 1. The foci are on the y-axis, so the slopes are \displaystyle \frac{a}{b} = \frac{4}{3}; with \displaystyle a = \frac{4}{3}b, \displaystyle \frac{25}{9}b^2 = 100 gives b = 6, a = 8.
\displaystyle \frac{y^2}{1} - \frac{x^2}{15} = 1. Here 2a = 2, a = 1, c = 4, b^2 = 16 - 1 = 15.
\displaystyle \frac{x^2}{16} - \frac{y^2}{36} = 1, foci \displaystyle (\pm 2\sqrt{13}, 0). Here \displaystyle \frac{b}{4} = \frac{3}{2}, so b = 6, and c^2 = 16 + 36 = 52.
Substituting: \displaystyle \frac{25}{16} - \frac{81}{16} \cdot \frac{1}{9} = \frac{25}{16} - \frac{9}{16} = 1. Distances: c^2 = 16 + 9 = 25, foci (\pm 5, 0); the distance to (5, 0) is \displaystyle \frac{9}{4}, to (-5, 0) is \displaystyle \sqrt{100 + \frac{81}{16}} = \frac{41}{4}, and \displaystyle \frac{41}{4} - \frac{9}{4} = 8 = 2a.
At x = 10: \displaystyle 5 - 2\sqrt{6} \approx 0.101. At x = 100: \displaystyle 50 - \sqrt{2499} \approx 0.0100. The curve has \displaystyle y = \frac{1}{2}\sqrt{x^2 - 4}; the gap is also \displaystyle \frac{2}{x + \sqrt{x^2 - 4}}, which shrinks about ten times when x grows ten times.
Center (3, -1), vertices (0, -1) and (6, -1), foci (-2, -1) and (8, -1), asymptotes \displaystyle y + 1 = \pm\frac{4}{3}(x - 3). Here a = 3, b = 4, c = 5.
Center (4, 2), vertices (4, 7) and (4, -3), foci (4, 15) and (4, -11), asymptotes \displaystyle y - 2 = \pm\frac{5}{12}(x - 4). Opens up and down with a = 5, b = 12, c = 13.
\displaystyle \frac{(x + 1)^2}{1} - \frac{(y - 3)^2}{4} = 1: center (-1, 3), vertices (0, 3) and (-2, 3), foci \displaystyle (-1 \pm \sqrt{5}, 3), asymptotes y - 3 = \pm 2(x + 1). From 4(x^2 + 2x) - (y^2 - 6y) = 9, add 4 \cdot 1 = 4 and -1 \cdot 9 = -9 to get 4(x + 1)^2 - (y - 3)^2 = 4, then divide by 4; c^2 = 1 + 4 = 5.
\displaystyle \frac{(y + 2)^2}{9} - \frac{(x - 1)^2}{16} = 1: center (1, -2), vertices (1, 1) and (1, -5), foci (1, 3) and (1, -7), asymptotes \displaystyle y + 2 = \pm\frac{3}{4}(x - 1). From 16(y^2 + 4y) - 9(x^2 - 2x) = 89, add 64 and -9 to get 16(y + 2)^2 - 9(x - 1)^2 = 144.
Corners (-10, 4), (2, 4), (2, -2), (-10, -2); asymptotes \displaystyle y - 1 = \pm\frac{1}{2}(x + 4). The center is (-4, 1) with a = 6 across and b = 3 up and down.
\displaystyle \frac{(x - 3)^2}{16} - \frac{(y - 3)^2}{9} = 1. The center is the midpoint (3, 3); the vertices are 4 from it and the foci 5, so b^2 = 25 - 16 = 9.
\displaystyle \frac{(y + 1)^2}{9} - \frac{(x + 2)^2}{4} = 1, foci \displaystyle (-2, -1 + \sqrt{13}) and \displaystyle (-2, -1 - \sqrt{13}). The vertex is 3 above the center, so a = 3 and the branches open up and down; the slope \displaystyle \frac{a}{b} = \frac{3}{2} gives b = 2, and c^2 = 9 + 4 = 13.
\displaystyle \frac{(y + 1)^2}{16} - \frac{(x - 1)^2}{9} = 1. The center is (1, -1) and c = 5; 2a = 8 gives a = 4, and b^2 = 25 - 16 = 9.
(a) \displaystyle \frac{x^2}{462{,}400} - \frac{y^2}{537{,}600} = 1, the branch around the station that heard the bang first: 2a = 4 \cdot 340 = 1360, a = 680, c = 1000, b^2 = 1{,}000{,}000 - 462{,}400. (b) c - a = 320 meters, at the vertex.
(a) \displaystyle \frac{x^2}{8100} - \frac{y^2}{14{,}400} = 1: 2a = 0.0006 \cdot 300{,}000 = 180, a = 90, c = 150, b^2 = 22{,}500 - 8100. (b) 60 kilometers: the ship is at a vertex, c - a from the nearer station.
For any point P, the triangle inequality gives |PF_1 - PF_2| \le F_1F_2, with equality only on the two rays of the line through the foci that start at the foci and point away from each other. So 2a = F_1F_2 gives those two rays, 2a > F_1F_2 gives no points, and only 2a < F_1F_2 gives a curve with two branches.
On \displaystyle y = \frac{b}{a}x the left side is \displaystyle \frac{x^2}{a^2} - \frac{x^2}{a^2} = 0 \ne 1, and the same for \displaystyle y = -\frac{b}{a}x. And \displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 means \displaystyle y^2 = \frac{b^2}{a^2}x^2, that is \displaystyle y = \frac{b}{a}x or \displaystyle y = -\frac{b}{a}x.
The value c = 5 is right, from c^2 = 9 + 16 = 25. But a^2 is the denominator under the positive term, 9, so the branches open up and down: vertices (0, \pm 3), foci (0, \pm 5), asymptotes \displaystyle y = \pm\frac{3}{4}x. Check at (0, 3): distances 2 and 8 differ by 6 = 2a.
Since \displaystyle \frac{y^2}{b^2} \ge 0, every point has \displaystyle \frac{x^2}{a^2} = 1 + \frac{y^2}{b^2} \ge 1, so |x| \ge a. No point lies in the strip -a < x < a, so the points with x \ge a and those with x \le -a form two separate pieces. Every x with |x| \ge a gives a value of y, and |y| grows without bound as |x| does, so neither branch is bounded, unlike an ellipse.