Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a hyperbola;
(2) determine the standard form of equation of a hyperbola;
(3) graph a hyperbola in a rectangular coordinate system; and (4) solve situational problems involving conic sections (hyperbolas).
Lesson Outline
(1) Definition of a hyperbola
(2) Derivation of the standard equation of a hyperbola (3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas Introduction
Just like ellipse, a hyperbola is one of the conic sections that most students have not encountered formally before. Its graph consists of two unbounded branches which extend in opposite directions. It is a misconception that each branch is a parabola. This is not true, as parabolas and hyperbolas have very different features. An application of hyperbolas in basic location and navigation schemes are presented in an example and some exercises.
1.4.1. Definition and Equation of a Hyperbola
Consider the points F1(−5, 0) and F2(5, 0) as shown in Figure 1.42. What is the absolute value of the difference of the distances of A(3.75, −3) from F1 and from
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F2? How about the absolute value of the difference of the distances of B −5,163 from F1 and from F2?
|AF1− AF2| = |9.25 − 3.25| = 6
|BF1− BF2| =
16 3 − 34
3
= 6
There are other points P such that |P F1 − P F2| = 6. The collection of all such points forms a shape called a hyperbola, which consists of two disjoint branches.
For points P on the left branch, P F2− P F1 = 6; for those on the right branch, P F1− P F2 = 6.
Figure 1.42
Figure 1.43
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Let F1 and F2 be two distinct points. The set of all points P , whose distances from F1 and from F2 differ by a certain constant, is called a hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.43, given are two points on the x-axis, F1(−c, 0) and F2(c, 0), the foci, both c units away from their midpoint (0, 0). This midpoint is the center of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute value of the difference of the distances of P from F1 and F2, be 2a (the coefficient 2 will make computations simpler). Thus, |P F1− P F2| = 2a, and so
Algebraic manipulations allow us to rewrite this into the much simpler x2 that P is not on the x-axis, so 4P F1F2 is formed. From the triangle inequality, F1F2+ P F2 > P F1. Thus, 2c > P F1− P F2 = 2a, so c > a.
We collect here the features of the graph of a hyperbola with standard equa-tion
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Figure 1.44 Figure 1.45
(1) center : origin (0, 0)
(2) foci : F1(−c, 0) and F2(c, 0)
• Each focus is c units away from the center.
• For any point on the hyperbola, the absolute value of the difference of its distances from the foci is 2a.
(3) vertices: V1(−a, 0) and V2(a, 0)
• The vertices are points on the hyperbola, collinear with the center and foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = bax and y = −bax, the lines `1 and `2 in Figure 1.45
• The asymptotes of the hyperbola are two lines passing through the cen-ter which serve as a guide in graphing the hyperbola: each branch of the hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. (We need the concept of limits from calculus to explain this.)
• An aid in determining the equations of the asymptotes: in the standard equation, replace 1 by 0, and in the resulting equation xa22−yb22 = 0, solve for y.
• To help us sketch the asymptotes, we point out that the asymptotes
`1 and `2 are the extended diagonals of the auxiliary rectangle drawn in Figure 1.45. This rectangle has sides 2a and 2b with its diagonals intersecting at the center C. Two sides are congruent and parallel to the transverse axis V1V2. The other two sides are congruent and parallel
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to the conjugate axis, the segment shown which is perpendicular to the transverse axis at the center, and has length 2b.
Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola with equation
x2 9 − y2
7 = 1.
Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
Solution. With a2 = 9 and b2 = 7, we have
The graph is shown at the right. The conju-gate axis drawn has its endpoints b = √
7 ≈ 2.7 units above and below the center. 2
Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are F1(−5, 0) and F2(5, 0), such that for any point on it, the absolute value of the difference of its distances from the foci is 6. See Figure 1.42.
Solution. We have 2a = 6 and c = 5, so a = 3 and b = √
c2− a2 = 4. The hyperbola then has equation x2
9 − y2
16 = 1. 2
1.4.2. More Properties of Hyperbolas
The hyperbolas we considered so far are “horizontal” and have the origin as their centers. Some hyperbolas have their foci aligned vertically, and some have centers not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are not shown anymore.
In all four cases below, we let c = √
a2+ b2. The foci F1 and F2 are c units away from the center C. The vertices V1 and V2 are a units away from the center.
The transverse axis V1V2 has length 2a. The conjugate axis has length 2b and is perpendicular to the transverse axis. The transverse and conjugate axes bisect each other at their intersection point, C. Each branch of a hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends.
The equations of the asymptotes can be determined by replacing 1 in the standard equation by 0. The asymptotes can be drawn as the extended diagonals of the auxiliary rectangle determined by the transverse and conjugate axes. Recall that,
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for any point on the hyperbola, the absolute value of the difference of its distances from the foci is 2a.
Center Corresponding Hyperbola
(0, 0)
x2 a2 − y2
b2 = 1 y2
a2 −x2 b2 = 1
(h, k)
(x − h)2
a2 − (y − k)2
b2 = 1 (y − k)2
a2 − (x − h)2 b2 = 1
transverse axis: horizontal transverse axis: vertical conjugate axis: vertical conjugate axis: horizontal
In the standard equation, aside from being positive, there are no other re-strictions on a and b. In fact, a and b can even be equal. The orientation of the hyperbola is determined by the variable appearing in the first term (the positive term): the corresponding axis is where the two branches will open. For example, if the variable in the first term is x, the hyperbola is “horizontal”: the transverse
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axis is horizontal, and the branches open to the left and right in the direction of the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymp-totes of the hyperbola with the given equation. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rect-angle.
(1) (y + 2)2
25 − (x − 7)2
9 = 1
(2) 4x2− 5y2+ 32x + 30y = 1
Solution.√ (1) From a2 = 25 and b2 = 9, we have a = 5, b = 3, and c = a2+ b2 =√
34 ≈ 5.8. The hyperbola is vertical. To determine the asymp-totes, we write (y+2)25 2 −(x−7)9 2 = 0, which is equivalent to y + 2 = ±53(x − 7).
We can then solve this for y.
center: C(7, −2) foci: F1(7, −2 −√
34) ≈ (7, −7.8) and F2(7, −2 +√
34) ≈ (7, 3.8) vertices: V1(7, −7) and V2(7, 3)
asymptotes: y = 53x − 413 and y = −53x + 293
The conjugate axis drawn has its endpoints b = 3 units to the left and right of the center.
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(2) We first change the given equation to standard form.
4(x2+ 8x) − 5(y2− 6y) = 1
4(x2+ 8x + 16) − 5(y2− 6y + 9) = 1 + 4(16) − 5(9) 4(x + 4)2− 5(y − 3)2 = 20
(x + 4)2
5 − (y − 3)2
4 = 1
We have a = √
5 ≈ 2.2 and b = 2. Thus, c =√
a2+ b2 = 3. The hyperbola is horizontal. To determine the asymptotes, we write (x+4)5 2 − (y−3)4 2 = 0 which is equivalent to y − 3 = ±√2
5(x + 4), and solve for y.
center: C(−4, 3)
foci: F1(−7, 3) and F2(−1, 3) vertices: V1(−4 −√
5, 3) ≈ (−6.2, 3) and V2(−4 +√
5, 3) ≈ (−1.8, 3) asymptotes: y = √2
5x +√8
5 + 3 and y = −√2
5x − √8
5 + 3
The conjugate axis drawn has its endpoints b = 2 units above and below the center.
Example 1.4.4. The foci of a hyperbola are (−5, −3) and (9, −3). For any point on the hyperbola, the absolute value of the difference of its of its distances from the foci is 10. Find the standard equation of the hyperbola.
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Solution. The midpoint (2, −3) of the foci is the center of the hyperbola. Each focus is c = 7 units away from the center. From the given difference, 2a = 10 so a = 5. Also, b2 = c2− a2 = 24. The hyperbola is horizontal (because the foci are horizontally aligned), so the equation is
(x − 2)2
25 − (y + 3)2
24 = 1. 2
Example 1.4.5. A hyperbola has vertices (−4, −5) and (−4, 9), and one of its foci is (−4, 2 −√
65). Find its standard equation.
Solution. The midpoint (−4, 2) of the vertices is the center of the hyperbola, which is vertical (because the vertices are vertically aligned). Each vertex is a = 7 units away from the center. The given focus is c = √
65 units away from the center. Thus, b2 = c2− a2 = 16, and the standard equation is
(y − 2)2
49 − (x + 4)2
16 = 1. 2
1.4.3. Situational Problems Involving Hyperbolas Let us now give an example on an application of hyperbolas.
Example 1.4.6. An explosion was heard by two stations 1200 m apart, located at F1(−600, 0) and F2(600, 0). If the explosion was heard in F1two seconds before it was heard in F2, identify the possible locations of the explosion. Use 340 m/s as the speed of sound.
Solution. Using the given speed of sound, we can deduce that the sound traveled 340(2) = 680 m farther in reaching F2 than in reaching F1. This is then the difference of the distances of the explosion from the two stations. Thus, the explosion is on a hyperbola with foci are F1 and F2, on the branch closer to F1.
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We have c = 600 and 2a = 680, so a = 340 and b2 = c2 − a2 = 244400.
The explosion could therefore be anywhere on the left branch of the hyperbola
x2
115600 −244400y2 = 1. 2
More Solved Examples
1. Determine the foci, vertices, and asymptotes of the hyperbola with equation x2
16 − y2
33 = 1. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
Solution: The hyperbola is horizontal.
a2 = 16 ⇒ a = 4, The conjugate axis has endpoints (0, −√
33) and (0,√
33). See Figure 1.46.
Figure 1.46
2. Find the standard equation of the hyperbola whose foci are F1(0, −10) and F2(0, 10), such that for any point on it, the absolute value of the difference of its distances from the foci is 12.
Solution: The hyperbola is vertical and has center at (0, 0). We have 2a = 12, so a = 6; also, c = 10. Then b = √
102− 62 = 8. The equation is y2 36−x2
64 = 1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and in-clude these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
3. (y + 6)2
25 − (x − 4)2 39 = 1
Solution: The hyperbola is vertical.
a2 = 25 ⇒ a = 5, b2 = 39 ⇒ b =√
39, c =√
25 + 39 = 8
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center: (4, −6)
foci: F1(4, −14), F2(4, 2) vertices: V1(4, −11), V2(4, −1) asymptotes: (y + 6)2
25 − (x − 4)2 39 = 0
⇔ y + 6 = ± 5
√39(x − 4) The conjugate axis has endpoints b =√
39 units to the left and to the right of the center. See Figure 1.47.
Figure 1.47
4. 9x2+ 126x − 16y2− 96y + 153 = 0 Solution:
9x2+ 126x − 16y2− 96y = −153
9(x2+ 14x + 49) − 16(y2+ 6y + 9) = −153 + 9(49) − 16(9) 9(x + 7)2− 16(y + 3)2 = 144
(x + 7)2
16 − (y + 3)2
9 = 1
The hyperbola is horizontal.
a2 = 16 ⇒ a = 4, b2 = 9 ⇒ b = 3, c = √
16 + 9 = 5
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center: (−7, −3)
foci: F1(−12, −3), F2(−2, −3) vertices: V1(−11, −3), V2(−3, −3) asymptotes: (x + 7)2
16 − (y + 3)2
9 = 0 ⇔ y + 3 = ±3
4(x + 7) The conjugate axis have endpoints (−7, −6) and (−7, 0). See Figure 1.48.
Figure 1.48
5. The foci of a hyperbola are (−17, −3) and (3, −3). For any point on the hyperbola, the absolute value of the difference of its distances from the foci is 14. Find the standard equation of the hyperbola.
Solution: The hyperbola is horizontal with center at the midpoint (−7, −3) of the foci. Also, 2a = 14 so a = 7 while c = 10. Then b2 = 102− 72 = 51. The equation is (x + 7)2
49 −(y + 3)2 51 = 1.
6. The auxiliary rectangle of a hyperbola has vertices (−24, −15), (−24, 9), (10, 9), and (10, −15). Find the equation of the hyperbola if its conjugate axis is hor-izontal.
Solution: The hyperbola is vertical. Using the auxiliary rectangle’s dimen-sions, we see that the length of the transverse axis is 2a = 24 while the length of the conjugate axis is 2b = 34. Thus, a = 12 and b = 17. The hyperbola’s vertices are the midpoints (−7, −15) and (−7, 9) of the bottom
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and top sides, respectively, of the auxiliary rectangle. Then the hyperbola’s center is (−7, −3), which is the midpoint of the vertices. The equation is
(y + 3)2
144 − (x + 7)2 289 = 1.
7. Two LORAN (long range navigation) stations A and B are situated along a straight shore, where A is 200 miles west of B. These stations transmit radio signals at a speed 186 miles per millisecond. The captain of a ship travelling on the open sea intends to enter a harbor that is located 40 miles east of station A.
Due to the its location, the harbor experiences a time difference in receiving the signals from both stations. The captain navigates the ship into the harbor by following a path where the ship experiences the same time difference as the harbor.
(a) What time difference between station signals should the captain be look-ing for in order the ship to make a successful entry into the harbor?
(b) If the desired time difference is achieved, determine the location of the ship if it is 75 miles offshore.
Solution:
(a) Let H represent the harbor on the shoreline. Note that BH −AH = 160−
40 = 120. The time difference on the harbor is given by 120÷186 ≈ 0.645 milliseconds. This is the time difference needed to be maintained in order to for the ship to enter the harbor.
(b) Situate the stations A and B on the Cartesian plane so that A (−100, 0) and B (100, 0). Let P represent the ship on the sea, which has coordinates (h, 75). Since P B − P A = 120, then it should follow that h < 0. More-over, P should lie on the left branch of the hyperbola whose equation is given by
This means that the ship is around 17.76 miles to the east of station A.
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Supplementary Problems 1.4
Give the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph and the auxiliary rectangle, then include these points and lines.
1. x2 100 − y2
81 = 1 2. y − x = 1
y + x
3. 4x2− 15(y − 5)2 = 60
4. (y − 6)2
64 −(x − 8)2 36 = 1 5. 9y2 + 54y − 6x2− 36x − 27 = 0
6. 16x2+ 64x − 105y2+ 840y − 3296 = 0
Find the standard equation of the hyperbola which satisfies the given conditions.
7. foci (−7, −17) and (−7, 17), the absolute value of the difference of the distances of any point from the foci is 24
8. foci (−3, −2) and (15, −2), a vertex at (9, −2)
9. center (−10, −4), one corner of auxiliary rectangle at (−1, 12), with horizontal transverse axis
10. asymptotes y = 713 − 43x and y = 43x −173 and a vertex at (17, 9) 11. asymptotes y = −125x +193 and y = 125x +293 and a focus at (−4, −5)
12. horizontal conjugate axis, one corner of auxiliary rectangle at (3, 8), and an asymptote 4x + 3y = 12
13. two corners of auxiliary rectangle at (2, 3) and (16, −1), and horizontal trans-verse axis
14. Two radio stations are located 150 miles apart, where station A is west of sta-tion B. Radio signals are being transmitted simultaneously by both stasta-tions, travelling at a rate of 0.2 miles/µsec. A plane travelling at 60 miles above ground level has just passed by station B and is headed towards the other station. If the signal from B arrives at the plane 480 µsec before the signal sent from A, determine the location of the plane.
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