Hyperbolas
by CHED on June 16, 2017
lesson duration of 5 minutes
under Precalculus
generated on June 16, 2017 at 06:32 pm
Tags: Analytic Geometry
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Generated: Jun 17,2017 02:32 AM
Hyperbolas
( 5 mins )
Written By: CHED on June 30, 2016
Subjects: Precalculus
Tags: Analytic Geometry
Resources
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Content Standard
The learners demonstrate an understanding of key concepts of conic sections and systems of nonlinear equations
Performance Standard
The learners shall be able to model situations appropriately and solve problems accurately using conic sections and
systems of nonlinear equations
Learning Competencies
The learners will graph a hyperbola in a rectangular coordinate system
The learners will determine the standard form of equation of a hyperbola
The learners will define a hyperbola
Definition and Equation of a Hyperbola 1 mins
A hyperbola is one of the conic sections that most students have not encountered formally before, unlike circles and
parabolas. 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.
Consider the points F1(–5, 0) and F2(5, 0) as shown in Figure 1.23. What is the absolute value of the difference of the
distances of A(3.75,–3) from F1 and from F2? How about the absolute value of the difference of the distances of B " –5,
16 3 # from F1 and from F2?
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|AF1 – AF2| = |9.25 – 3.25| = 6
|BF1 – BF2| = |16/3 – 34/3| = 6
There are other points P such that |PF1 ? PF2| = 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, PF2 – PF1 = 6; for those on the right branch,
PF1 – PF2 = 6.
Let F1 and F2 be two distinct points. The set of all points P, whose distances from F1 and from F2 dffer by a certain
constant, is called a hyperbola.
hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.24, 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,
|PF1 – PF2| = 2a, and so
Algebraic manipulations allow us to rewrite this into the much simpler
When we let
, we assumed c > a. To see why this is true, suppose that
P is closer to F2, so PF1 – PF2 = 2a. Refer to Figure 1.24. Suppose also 46 that P is not on the x-axis, ?PF1F2 is
formed. From the triangle inequality, F1F2 + PF2 > PF1. Thus, 2c > PF1 – PF2 = 2a, so c > a.
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Teaching Notes
Notes
If it is assumed that P is closer to F1, then the same equation will be obtained because of symmetry.
We collect here the features of the graph of a hyperbola with standard equation.
x2/a2 – y2/b2 = 1.
(1) center:
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.
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(4) asymptotes: y = (b/a)x and y = –(b/a) x, the lines
1
and
2
in Figure 1.26
• The asymptotes of the hyperbola are two lines passing through the center 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 x2/a2 – y2/b2 = 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 1.26. 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 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.
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.
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Solution. With a2 = 9 and b2 = 7, we have a = 3, b = ?7, and
4.
=
foci: F1(?4, 0) and F2(4, 0)
vertices: V1(–3, 0) and V2(3, 0)
asymptotes: y = (?7/3) x and y = ( –?7/3) x
The graph is shown at the right. The conjugate axis drawn has its endpoints b = ?7 ? 2.7 units above and below the
center.
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 1.23.
Solution. We have 2a = 6 and c = 5, so a = 3 and
has equation x2 /9 – y2 /16 = 1.
= 4. The hyperbola then
Seatwork/Homework 1 mins
1. Determine foci, vertices, and asymptotes of the hyperbola with equation
x2/16 – y2/20 = 1.
Sketch the graph, and include these points and lines, along with the auxiliary rectangle.
Answer: foci F1(–6, 0) and F2(6, 0), vertices V1(–4, 0) and V2(4, 0), asymptotes y = (?5/2) x and y =( – ?5/2) x
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2. Find the equation in standard form of the hyperbola whose foci are F1(–4?2, 0) and F2(4?2, 0), such that for any
point on it, the absolute value of the difference of its distances from the foci is 8.
Answer: x2/16 – y2/16 = 1
More Properties of Hyperbolas 0 mins
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.
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In all four cases above, we let
. 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, for any point on the hyperbola, the absolute value of the
difference of its distances from the foci is 2a.
In the standard equation, aside from being positive, there are no other restrictions 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 axis is horizontal, and the branches open to the left and right in the direction
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of the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given
equation. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary
rectangle.
(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
= ?34 ? 5.8. The hyperbola is vertical. To determine the asymptotes, we
write (y+2)2/25 – (x–7)2/9 = 0, which is equivalent to y+2 = ±5/3 (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 = 5/3x – 41/3 and y = –5/3x + 29/3
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.
We have a = ?5 ? 2.2 and b = 2. Thus,
= 3. The hyperbola is horizontal.
To determine the asymptotes, we write (x+4)2/5 – (y–3)2/4 = 0 which is equivalent to y –3 = ± 2 p5 (x + 4), and solve for
y.
center: C(–4, 3)
foci: F1(–7, 3) and F2(–1, 3)
vertices: V1(–4 – p5, 3) ? (–6.2, 3) and V2(–4 + p5, 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.
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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.
Solution.
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.
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.
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.
Seatwork/Homework 1 mins
We now give an example on an application of hyperbolas
Example 1.4.6.
1.4.6. An explosion is heard by two stations 1200 m apart, located at F1(–600, 0) and F2(600, 0). If the
explosion was heard in F1 two 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 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 – y2 / 244400 = 1.
Situational Problems Involving Hyperbolas 0 mins
We now give an example on an application of hyperbolas.
Example 1.4.6.
1.4.6. An explosion is heard by two stations 1200 m apart, located at F1(?600, 0) and F2(600, 0). If the
explosion was heard in F1 two 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 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.
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 ? y2 244400 = 1.
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Seatwork/Homework 1 mins
*1. Two stations, located at M(–1.5, 0) and N(1.5, 0) (units are in km), simultaneously send sound signals to a ship,
with the signal traveling at the speed of 0.33 km/s. If the signal from N was received by the ship four seconds before
the signal it received from M, find the equation of the curve containing the possible location of the ship.
Answer: x2 / 0.4356 – y2 / 1.8144 = 1 (right branch)
Exercises 1 mins
1. Give the coordinates of the center, foci, vertices, and the asymptotes of the hyperbola with the given equation.
Sketch the graph, and include these points and lines.
Answer:
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2. Find the standard equation of the hyperbola which satisfies the given conditions.
(a) foci (–4,?3) and (–4, 13), the absolute value of the difference of the distances of any point from the foci is 14
Answer: (y – 5)2 / 49 – (x + 4)2 / 15 = 1
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(b) vertices (–2, 8) and (8, 8), a focus (12, 8)
Answer: (x – 3)2 / 25 – (y – 8)2 / 56 = 1
(c) center (–6, 9), a vertex (–6, 15), conjugate axis of length 12
Answer: (y – 9)2 / 25 – (x + 6)2 / 36 = 1
(d) asymptotes y = 4 / 3x + 1 / 3 and y = –4 / 3x + 41 / 3 , a vertex (1–, 7)
Answer: (x – 5)2 / 36 – (y – 7)2 / 64 = 1
Solution. The asymptotes intersect at (5, 7). This is the center. The distance of the given vertex from the center is a =
6. This vertex and center are aligned horizontally, so the hyperbola has equation of the form (x–h)2 / a2 – (y–k)2 / b2 =
1. The asymptotes consequently have the form y ? k = ±b a (x ? h), and thus, have slopes ±b a . From the given
asymptotes, b / a = 4 / 3 . Since a = 6, then b = 8. The standard equation is then
(x ? 5)2 / 36 ? (y ? 7)2 / 64 = 1.
(e) asymptotes y = 1 / 3x + 5 / 3 and y = –1 / 3x + 7 / 3, a focus (1, 12)
Answer: (y – 2)2 10 – (x – 1)2 90 = 1
Solution.
Solution. The asymptotes intersect at (1, 2). This is the center. The distance of the given focus from the center is c =
10. This focus and center are aligned vertically, so the hyperbola has equation of the form (y – k)2 / a2 – (x – h)2 / b2 =
1. The asymptotes consequently have the form y – k = ±a b (x?h), and thus, have slopes ±a b . From the given
asymptotes, a b = 1 / 3 , so b = 3a.
c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2
Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is
(y – 2)2 / 10 – (x – 1)2 / 90 = 1.
3. Two control towers are located at points Q(–500, 0) and R(500, 0), on a straight shore where the x-axis runs through
(all distances are in meters). At the same moment, both towers sent a radio signal to a ship out at sea, each traveling
at 300 m/µs. The ship received the signal from Q 3 µs (microseconds) before the message from R.
(a) Find the equation of the curve containing the possible location of the ship.
Answer: x2 / 202500 – y2 / 47500 = 1 (left branch)
(b) Find the coordinates (rounded off to two decimal places) of the ship if it is 200 m from the shore (y = 200).
Answer: (– 610.76, 200)
Solution. Since the time delay between the two signals is 3 µs, then the difference between the distances traveled by
the two signals is 300 ·3 = 900 m. The ship is then on a hyperbola, consisting of points whose distances from Q and R
(the foci) differ by 2a = 900. With a = 450 and c = 500 (the distance of each focus from the center, the origin), we have
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b2 = c2–a2 = 5002–4502 = 47500. Since a2 = 202500, the hyperbola then has equation x2 / 202500 – y2 / 47500 = 1.
Since the signal from Q was received first, the ship is closer to Q than R, so the ship is on the left branch of this
hyperbola. Using y = 200, we then solve x2 / 202500 – 2002 / 47500 = 1 for x < 0 (left branch), and we get x ? –610.76.
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