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Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one of the foci is the star around which it revolves. The planet is closest to the star when it is at one vertex. It is farthest from the star when it is at the other vertex. Suppose the closest and farthest distances of the planet from this star, are 420 million kilometers and 580 million kilometers, respectively. Find the equation of the ellipse, in standard form, with center at the origin and the star at the x-axis. Assume all units are in millions of kilometers.
Solution. In the figure above, the orbit is drawn as a horizontal ellipse with center at the origin. From the planet’s distances from the star, at its closest and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000 (million kilometers), so a = 500. If we place the star at the positive x-axis, then it is c = 500 − 420 = 80 units away from the center. Therefore, we get b^2 = a^2 − c^2 = 500^2 − 802 = 243600. The equation then is
x^2 250000
y^2 243600
The star could have been placed on the negative x-axis, and the answer would still be the same.
Seatwork/Homework 1.3.
1. The arch of a bridge is in the shape of a semiellipse, with its major axis at the
water level. Suppose the arch is 20 ft high in the middle, and 120 ft across its major axis. How high above the water level is the arch, at a point 20 ft from the center (horizontally)? Round off to 2 decimal places. Refer to Example 1.3.6. Answer: 18.86 ft
Exercises 1.
- Give the coordinates of the center, vertices, covertices, and foci of the ellipse with the given equation. Sketch the graph, and include these points.
(a)
x^2 169
y^2 25
(b)
x^2 144
y^2 169
(c) 4x^2 + 13y^2 = 52
(d)
(x + 7)^2 16
(y − 4)^2 25
(e) 9x^2 + 16y^2 + 72x − 96 y + 144 = 0 (f) 36x^2 + 20y^2 − 144 x + 120y − 396 = 0 Answer: Item Center Vertices Covertices Foci (a) (0, 0) (± 13 , 0) (0, ±5) (± 12 , 0) (b) (0, 0) (0, ±13) (± 12 , 0) (0, ±5)
(c) (0, 0) (±
(d) (− 7 , 4) (− 7 , −1) (− 11 , 4) (− 7 , 1) (− 7 , 9) (− 3 , 4) (− 7 , 7) (e) (− 4 , 3) (− 8 , 3) (− 4 , 0) (− 4 ±
(f) (2, −3) (2, −9) (2 ± 2
(a) (b) (c)
- A truck that is about to pass through the tunnel from the previous item is 10 ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel? Answer: Yes
- An orbit of a satellite around a planet is an ellipse, with the planet at one focus of this ellipse. The distance of the satellite from this star varies from 300 , 000 km to 500, 000 km, attained when the satellite is at each of the two vertices. Find the equation of this ellipse, if its center is at the origin, and the vertices are on the x-axis. Assume all units are in 100, 000 km. Answer: x
2 16 +^
y^2 15 = 1
- The orbit of a planet around a star is described by the equation 640 x,^2000 + y^2 630 , 000 = 1, where the star is at one focus, and all units are in millions of kilometers. The planet is closest and farthest from the star, when it is at the vertices. How far is the planet when it is closest to the sun? How far is the planet when it is farthest from the sun? Answer: 700 million km, 900 million km
Solution. The ellipse has center at the origin, and major axis on the x-axis. Since a^2 = 640 , 000, then a = 800, so the vertices are V 1 (− 800 , 0) and V 2 (800, 00). Since b^2 = 630, 000, then c =
a^2 − b^2 =
10 , 000 = 100. Sup- pose the star is at the focus at the right of the origin (this choice is arbitrary, since we could have chosen instead the focus on the left). Its location is then F (100, 0). The closest distance is then V 2 F = 700 (million kilometers) and the farthest distance is V 1 F = 900 (million kilometers).
- A big room is constructed so that the ceiling is a dome that is semielliptical in shape. If a person stands at one focus and speaks, the sound that is made bounces off the ceiling and gets reflected to the other focus. Thus, if two people stand at the foci (ignoring their heights), they will be able to hear each other. If the room is 34 m long and 8 m high, how far from the center should each of two people stand if they would like to whisper back and forth and hear each other? Answer: 15 m
Solution. We could put a coordinate system with the floor of the room on the x-axis, and the center of the room at the origin, as shown in the figures. The major axis has length 34, and the height of the room is half of the minor axis. The ellipse that contains the ceiling then has equation 17 x^22 + y
2 82 = 1. The distance of a focus from the center is c =
a^2 − b^2 =
172 − 82 = 15. Thus, the two people should stand 15 m away from the center.
- A whispering gallery has a semielliptical ceiling that is 9 m high and 30 m long. How high is the ceiling above the two foci? Answer: 5.4 m
Solution. As in the previous problem, put a coordinate system with the floor of the room on the x-axis, and the center of the room at the origin. The major axis has length 30, and half the minor axis is 9. The ellipse that contains the ceiling then has equation x
2 152 +^
y^2 92 = 1. The distance of a focus from the center is c =
a^2 − b^2 =
152 − 92 = 12. If we put x = 12 in the equation of the ellipse, we get 12
2 152 +^
y^2 92 = 1. Solving for^ y >^ 0 yields^ y^ =^
27 5 = 5.4. The height of the ceiling above each focus is 5.4 m.
Lesson 1.4. Hyperbolas
Time Frame: 3 one-hour sessions
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
Figure 1.23 Figure 1.
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 F 1 and F 2 , be 2a (the coefficient 2 will make computations simpler). Thus, |P F 1 − P F 2 | = 2a, and so ∣∣ ∣
(x + c)^2 + y^2 −
(x − c)^2 + y^2
∣ = 2a.
Algebraic manipulations allow us to rewrite this into the much simpler
x^2 a^2
y^2 b^2
= 1, where b =
c^2 − a^2.
When we let b =
c^2 − a^2 , we assumed c > a. To see why this is true, suppose that P is closer to F 2 , so P F 1 − P F 2 = 2a. Refer to Figure 1.22. Suppose also that P is not on the x-axis, so !P F 1 F 2 is formed. From the triangle inequality, F 1 F 2 + P F 2 > P F 1. Thus, 2c > P F 1 − P F 2 = 2a, so c > a. Now we present a derivation. For now, assume P is closer to F 2 so P F 1 > P F 2 , Teaching Notesand^ P F 1 −^ P F 2 = 2a. If it is assumed that P is closer to F 1 , then the same equation will be obtained because of symmetry.
P F 1 = 2a + P F 2 √ (x + c)^2 + y^2 = 2a +
(x − c)^2 + y^2 (√ (x + c)^2 + y^2
2 a +
(x − c)^2 + y^2
cx − a^2 = a
(x − c)^2 + y^2
(cx − a^2 )^2 =
a
(x − c)^2 + y^2
(c^2 − a^2 )x^2 − a^2 y^2 = a^2 (c^2 − a^2 ) b^2 x^2 − a^2 y^2 = a^2 b^2 by letting b =
c^2 − a^2 > 0 x^2 a^2
y^2 b^2
We collect here the features of the graph of a hyperbola with standard equa- tion x^2 a^2
y^2 b^2
Let c =
a^2 + b^2.
(1) center : origin (0, 0)
(2) foci: F 1 (−c, 0) and F 2 (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: V 1 (−a, 0) and V 2 (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 V 1 V 2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = (^) ab x and y = − (^) ab x, the lines 1 and 2 in Figure 1.
- 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 x
2 a^2 −^
y^2 b^2 = 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 1.24. 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 V 1 V 2. 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. Determine the foci, vertices, and asymptotes of the hyperbola with equation x^2 9
y^2 7
Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
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.
Center Corresponding Hyperbola
x^2 a^2
y^2 b^2
y^2 a^2
x^2 b^2
(h, k)
(x − h)^2 a^2
(y − k)^2 b^2
(y − k)^2 a^2
(x − h)^2 b^2
transverse axis: horizontal transverse axis: vertical conjugate axis: vertical conjugate axis: horizontal
In all four cases above, we let c =
a^2 + b^2. The foci F 1 and F 2 are c units away from the center C. The vertices V 1 and V 2 are a units away from the center. The transverse axis V 1 V 2 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 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 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
(2) 4x^2 − 5 y^2 + 32x + 30y = 1
Solution.√ (1) From a^2 = 25 and b^2 = 9, we have a = 5, b = 3, and c = a^2 + b^2 =
34 ≈ 5 .8. The hyperbola is vertical. To determine the asymp- totes, 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: F 1 (7, − 2 −
- ≈ (7, − 7 .8) and F 2 (7, −2 +
vertices: V 1 (7, −7) and V 2 (7, 3) asymptotes: y = 53 x − 413 and y = −^53 x + (^293)
The conjugate axis drawn has its endpoints b = 3 units to the left and right of 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.
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, b^2 = c^2 − a^2 = 24. The hyperbola is horizontal (because the foci are horizontally aligned), so the equation is
(x − 2)^2 25
(y + 3)^2 24
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, b^2 = c^2 − a^2 = 16, and the standard equation is
(y − 2)^2 49
(x + 4)^2 16
Seatwork/Homework 1.4.
- Give the coordinates of the center, foci, vertices, and asymptotes of the hy- perbola with equation 9x^2 − 4 y^2 − 90 x − 32 y = −305. Sketch the graph, and include these points and lines, along with the auxiliary rectangle. Answer: center C(5, −4), foci F 1 (5, − 4 − 2
- and F 2 (5, −4+
13), vertices V 1 (5, −10) and V 2 (5, 2), asymptotes y = −^32 x + 72 and y = 32 x − (^232)
- A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is (− 2 , 9). Find
its standard equation. Answer:
(x − 7)^2 36
(y − 9)^2 45
1.4.3. Situational Problems Involving Hyperbolas
We now give an example on an application of hyperbolas.
Example 1.4.6. An explosion is heard by two stations 1200 m apart, located at F 1 (− 600 , 0) and F 2 (600, 0). If the explosion was heard in F 1 two seconds before it was heard in F 2 , 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 F 2 than in reaching F 1. 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 F 1 and F 2 , on the branch closer to F 1.
Answer:
Item Center Vertices Foci
(a) (0, 0) (± 6 , 0) (± 10 , 0) (b) (0, 0) (0, ±5) (0, ±
(c) (1, 0) (− 1 , 0), (3, 0) (1 ± 2
(d) (− 3 , −2) (− 3 , − 2 ±
(e) (7, −4) (3, −4), (11, −4) (7 ± 2
(f) (− 3 , 5) (− 3 , 0), (− 3 , 10) (− 3 , −3), (− 3 , 13)
Item Asymptotes
(a) y = ±^43 x (b) y = ±^54 x (c) y = x − 1, y = −x + 1 (d) y = ±
3 2 x^ ±^3
3 2 −^2 (e) y = ±
3 2 x^ ∓^7
3 2 −^4 (f) y = ± √^539 x ± √^1539 + 5
(a) (b)
(c) (d)
(e) (f)
- Find the standard equation of the hyperbola which satisfies the given condi- tions.
(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
(b) vertices (− 2 , 8) and (8, 8), a focus (12, 8)
Answer:
(x − 3)^2 25
(y − 8)^2 56
(c) center (− 6 , 9), a vertex (− 6 , 15), conjugate axis of length 12
Answer:
(y − 9)^2 25
(x + 6)^2 36
(d) asymptotes y = 43 x + 13 and y = −^43 x + 413 , a vertex (− 1 , 7)
Answer:
(x − 5)^2 36
(y − 7)^2 64
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 a^2 −^
(y−k)^2 b^2 = 1. The asymptotes consequently have the form^ y^ −^ k^ =
Lesson 1.5. More Problems on Conic Sections
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) recognize the equation and important characteristics of the different types of conic sections; and
(2) solve situational problems involving conic sections.
Lesson Outline
(1) Conic sections with associated equations in general form
(2) Problems involving characteristics of various conic sections
(3) Solving situational problems involving conic sections
Introduction
Inspecting the equation can lead us to the right conic section for its graph, and set us on the right step towards analyzing it. We will also look at problems that use the properties of the different conic sections, allowing us to synthesize what has been covered so far.
1.5.1. Identifying the Conic Section by Inspection
The equation of a circle may be written in standard form
Ax^2 + Ay^2 + Cx + Dy + E = 0,
that is, the coefficients of x^2 and y^2 are the same. However, it does not follow that if the coefficients of x^2 and y^2 are the same, the graph is a circle.
General Equation Standard Equation graph (A) 2 x^2 + 2y^2 − 2 x + 6y + 5 = 0
x − (^12)
y + (^32)
= 0 point (B) x^2 + y^2 − 6 x − 8 y + 50 = 0 (x − 3)^2 + (y − 4)^2 = − 25 empty set
For a circle with equation (x − h)^2 + (y − k)^2 = r^2 , we have r^2 > 0. This is not the case for the standard equations of (A) and (B).
In (A), because the sum of two squares can only be 0 if and only if each square is 0, it follows that( x − 12 = 0 and y + 32 = 0. The graph is thus the single point 1 2 ,^ −
3 2
In (B), no real values of x and y can make the nonnegative left side equal to the negative right side. The graph is then the empty set.
Let us recall the general form of the equations of the other conic sections. We may write the equations of conic sections we discussed in the general form
Ax^2 + By^2 + Cx + Dy + E = 0.
Some terms may vanish, depending on the kind of conic section.
(1) Circle: both x^2 and y^2 appear, and their coefficients are the same
Ax^2 + Ay^2 + Cx + Dy + E = 0
Example: 18x^2 + 18y^2 − 24 x + 48y − 5 = 0 Degenerate cases: a point, and the empty set
(2) Parabola: exactly one of x^2 or y^2 appears
Ax^2 + Cx + Dy + E = 0 (D $= 0, opens upward or downward) By^2 + Cx + Dy + E = 0 (C $= 0, opens to the right or left)
Examples: 3 x^2 − 12 x + 2y + 26 = 0 (opens downward) − 2 y^2 + 3x + 12y − 15 = 0 (opens to the right)
(3) Ellipse: both x^2 and y^2 appear, and their coefficients A and B have the same sign and are unequal Examples: 2 x^2 + 5y^2 + 8x − 10 y − 7 = 0 (horizontal major axis) 4 x^2 + y^2 − 16 x − 6 y + 21 = 0 (vertical major axis) If A = B, we will classify the conic as a circle, instead of an ellipse. Degenerate cases: a point, and the empty set
(4) Hyperbola: both x^2 and y^2 appear, and their coefficients A and B have dif- ferent signs Examples: 5 x^2 − 3 y^2 − 20 x − 18 y − 22 = 0 (horizontal transverse axis) − 4 x^2 + y^2 + 24x + 4y − 36 = 0 (vertical transverse axis) Degenerate case: two intersecting lines The following examples will show the possible degenerate conic (a point, two intersecting lines, or the empty set) as the graph of an equation following a similar pattern as the non-degenerate cases.
(1) 4 x^2 + 9y^2 − 16 x + 18y + 25 = 0 =⇒
(x − 2)^2 32
(y + 1)^2 22
=⇒ one point: (2, −1)
(2) 4 x^2 + 9y^2 − 16 x + 18y + 61 = 0 =⇒
(x − 2)^2 32
(y + 1)^2 22
=⇒ empty set
