Ch. 11 Analytic Geometry
11.1 Conics
1 Know the Names of the Conics
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Name the conic.
1)
A) circle B) ellipse C) parabola D) hyperbola
2)
A) ellipse B) circle C) parabola D) hyperbola
3)
A) parabola B) circle C) ellipse D) hyperbola
Page 1
4)
A) hyperbola B) circle C) ellipse D) parabola
11.2 The Parabola
1 Analyze Parabolas with Vertex at the Origin
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Match the equation to its graph.
1) y2 = 10x
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 2
2) y2 = –14x
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 3
3) x2 = 15y
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 4
4) x2 = –8y
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Find the equation of the parabola described.
5) Focus at (–5
,
0); directrix the line x =5
A) y2 = –20x B) y2 = 20x C) y2 = –5x D) x2 = –20y
6) Focus at (0, –15); directrix the line y =15
A) x2 = –60y B) x2 = 60y C) y2 = –15x D) y2 = –60x
7) Focus at (6
,
0); vertex at (0, 0)
A) y2 = 24x B) x2 = 24y C) y2 = 6x D) x2 = 6y
8) Directrix the line y = 3; vertex at (0, 0)
A) x2 = –12y B) x = 3y2C) y2 = –12x D) y = –12x2
9) Focus at (5, 0); vertex at (0, 0)
A) y2 = 20x B) y = 20x2C) x2 = 20y D) x = 20y2
10) Vertex at (0, 0); axis of symmetry the x–axis; containing the point (3
,
1)
A) y2 = 1
3xB)y
2 = 1
12xC)x
2 = 1
3yD)x
2 = 1
12y
Page 5
Find an equation of the parabola described and state the two points that define the latus rectum.
11) Focus at (0, 4); directrix the line y = –4
A) x2 = 16y; latus rectum: (8, 4) and (–8, 4) B) y2 = 4x; latus rectum: (9, 2) and (–9, 2)
C) x2 = 16y; latus rectum: (4, 8) and (–4, 8) D) x2 = 4y; latus rectum: (2, 4) and (–2, 4)
Find the vertex, focus, and directrix of the parabola. Graph the equation.
12) x2 = –16y
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (0, 0)
focus: (0, –4)
directrix: y = 4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (0, 0)
focus: (0, 4)
directrix: y = –4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) vertex: (0, 0)
focus: (–4, 0)
directrix: x = 4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (0, 0)
focus: (4, 0)
directrix: x = –4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 6
13) y2 = 8x
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (0, 0)
focus: (2, 0)
directrix: x = –2
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (0, 0)
focus: (0, 2)
directrix: y = –2
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) vertex: (0, 0)
focus: (0, 2)
directrix: y = –2
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (0, 0)
focus: (–2, 0)
directrix: x = 2
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 7
Graph the equation.
14) y2 = 12x
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 8
15) y2 = –18x
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 9
16) x2 = 12y
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 10
17) x2 = –9y
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Write an equation for the parabola.
18)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(0, 0) (2, 1)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(0, 0) (2, 1)
A) x2 = 4y B) x2 = –4y C) y2 = 4x D) y2 = –4x
Page 11
19)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(0, 0)
(3, 6)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(0, 0)
(3, 6)
A) y2 = 12x B) x2 = –12y C) x2 = 12y D) y2 = –12x
2 Analyze Parabolas with Vertex at (h, k)
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Match the equation to the graph.
1) (y + 2)2 = 8(x + 2)
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 12
2) (y – 1)2 = –6(x + 2)
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 13
3) (x + 1)2 = 6(y – 2)
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 14
4) (x + 1)2 = –6(y + 1)
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Find the equation of the parabola described.
5) Vertex at (3
,
5); focus at (4
,
5)
A) (y – 5)2 = 4(x – 3) B) (y – 5)2 = –4(x – 3)
C) (x – 5)2 = –4(y – 5) D) (x – 5)2 = 4(y – 5)
6) Vertex at (1
,
8); focus at (1
,
7)
A) (x – 1)2 = –4(y – 8) B) (x – 1)2 = 4(y – 8)
C) (y – 8)2 = –24(x – 1) D) (y – 8)2 = 24(x – 1)
7) Vertex at (5
,
–3); focus at (5
,
–7)
A) (x – 5)2 = –16(y + 3) B) (x – 5)2 = 16(y + 3)
C) (y – 3)2 = 8(x + 5) D) (y – 3)2 = –8(x + 5)
8) Vertex at (3
,
–1); focus at (8
,
–1)
A) (y + 1)2 = 20(x – 3) B) (y + 1)2 = –20(x – 3)
C) (x + 3)2 = –28(y – 1) D) (x + 3)2 = 28(y – 1)
9) Focus at (–3
,
5); directrix the line y =1
A) (x + 3)2 = 8(y – 3) B) (x – 3)2 = 8(y – 3) C) (x + 3)2 = 8(y + 3) D) (x – 3)2 = 8(y + 3)
Page 15
Find the vertex, focus, and directrix of the parabola with the given equation.
10) (y – 2)2 = 4(x + 4)
A) vertex: (–4
,
2)
focus: (–3, 2)
directrix: x = –5
B) vertex: (4
,
–2)
focus: (5, –2)
directrix: x = 3
C) vertex: (2
,
–4)
focus: (3, –4)
directrix: x = 1
D) vertex: (–4
,
2)
focus: (–5, 2)
directrix: x = –3
11) (y – 1)2 = –12(x – 2)
A) vertex: (2
,
1)
focus: (–1, 1)
directrix: x = 5
B) vertex: (–2
,
–1)
focus: (–5, –1)
directrix: x = 1
C) vertex: (1
,
2)
focus: (–2, 2)
directrix: x = 4
D) vertex: (2
,
1)
focus: (5, 1)
directrix: x = –1
12) (x – 3)2 = 4(y – 1)
A) vertex: (3
,
1)
focus: (3, 2)
directrix: y = 0
B) vertex: (–3
,
–1)
focus: (–3, 0)
directrix: y = –2
C) vertex: (1
,
3)
focus: (1, 4)
directrix: y = 2
D) vertex: (3
,
1)
focus: (3, 0)
directrix: x = 2
13) (x – 1)2 = –4(y – 2)
A) vertex: (1
,
2)
focus: (1, 1)
directrix: y = 3
B) vertex: (–1
,
–2)
focus: (–1, –3)
directrix: y = –1
C) vertex: (2
,
1)
focus: (2, 0)
directrix: y = 2
D) vertex: (1
,
2)
focus: (1, 3)
directrix: x = 1
Find the vertex, focus, and directrix of the parabola. Graph the equation.
14) (y – 3)2 = 8(x + 1)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (–1
,
3)
focus: (1, 3)
directrix: x = –3
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (–3
,
1)
focus: (–1, 1)
directrix: x = –5
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 16
C) vertex: (–1
,
3)
focus: (–1, 5)
directrix: y = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (1
,
–3)
focus: (1, –1)
directrix: y = –5
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
15) (x + 3)2 =(y – 3)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (–3
,
3)
focus: (–3, 3.25)
directrix: y = 2.75
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (3
,
–3)
focus: (3, –2.75)
directrix: y = –3.25
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 17
C) vertex: (–3
,
3)
focus: (–2.75, 3)
directrix: x = –3.25
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (3
,
–3)
focus: (3.25, –3)
directrix: x = 2.75
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
16) x2 – 8x = 4y – 32
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (4
,
4)
focus: (4, 5)
directrix: y = 3
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (4
,
4)
focus: (4, 3)
directrix: y = 5
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 18
C) vertex: (4
,
4)
focus: (5, 4)
directrix: x = 3
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (4
,
4)
focus: (3, 4)
directrix: x = 5
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
17) y2 + 14y = 12x – 13
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (–3
,
–7)
focus: (0, –7)
directrix: x = –6
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (–3
,
–7)
focus: (–6, –7)
directrix: x = 0
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 19
C) vertex: (–3
,
–7)
focus: (–3, –4)
directrix: y = –10
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (–3
,
–7)
focus: (–3, –10)
directrix: y = –4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Graph the equation.
18) (y – 2)2 = 8(x – 1)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 20
19) (y + 1)2 = –5(x + 2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 21
20) (x + 2)2 = 5(y + 1)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 22
21) (x + 1)2 = –6(y + 1)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Write an equation for the parabola.
22)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
points: (1, –4), (3, 0)
A) (y + 4)2 = 8(x – 1) B) (y + 1)2 = 8(x – 4) C) (x + 4)2 = 8(y – 1) D) (x – 4)2 = 8(y + 1)
Page 23
23)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
points: (–4, 3), (0, 5)
A) (x + 4)2 = 8(y – 3) B) (x + 3)2 = 8(y – 4) C) (y + 4)2 = 8(x – 3) D) (x – 3)2 = 8(y + 4)
24)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
points: (3, –2), (7, –4)
A) (x – 3)2 = –8(y + 2) B) (x – 2)2 = –8(y + 3)
C) (y – 3)2 = –8(x + 2) D) (x + 2)2 = 8(y – 3)
3 Solve Applied Problems Involving Parabolas
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
1) A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches
across at its opening and is 4 feet deep, where will the light be concentrated?
A) 0.8 in. from the vertex B) 9 in. from the vertex
C) 0.7 in. from the vertex D) 0.3 in. from the vertex
2) A searchlight is shaped like a paraboloid of revolution. If the light source is located 3 feet from the base
along the axis of symmetry and the opening is 10 feet across, how deep should the searchlight be?
A) 2.1 ft B) 0.5 ft C) 8.3 ft D) 6.3 ft
3) A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 140 feet and a maximum
height of 40 feet. Find the height of the arch at 5 feet from its center.
A) 39.8 ft B) 0.1 ft C) 0.8 ft D) 1.1 ft
Page 24
4) A reflecting telescope has a mirror shaped like a paraboloid of revolution. If the distance of the vertex to
the focus is 28 feet and the distance across the top of the mirror is 52 inches, how deep is the mirror in the
center?
A) 169
336 in. B) 169
28 in. C) 169
4032 in. D) 98
13 in.
5) An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section,
cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top
of a second tower. The towers are both 6.25 inches tall and stand 50 inches apart. Find the vertical distance
from the roadway to the cable at a point on the road 7.5 inches from the lowest point of the cable.
A) 0.56 in. B) 2.25 in. C) 0.76 in. D) 0.36 in.
6) An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section,
cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top
of a second tower. The towers are both 16 inches tall and stand 80 inches apart. At some point along the
road from the lowest point of the cable, the cable is 2.56 inches above the roadway. Find the distance
between that point and the base of the nearest tower.
A) 24 in. B) 15.8 in. C) 24.2 in. D) 16.2 in.
7) An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section,
cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top
of a second tower. The towers stand 80 inches apart. At a point between the towers and 20 inches along the
road from the base of one tower, the cable is 4 inches above the roadway. Find the height of the towers.
A) 16 in. B) 16.5 in. C) 15.5 in. D) 18 in.
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
8) A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike
the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 8 feet
across at its opening and is 2 feet deep at its center, at what position should the receiver be placed?
9) A sealed–beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the
focus, is 3 centimeters from the vertex. If the depth is to be 6 centimeters, what is the diameter of the
headlight at its opening?
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
10) A spotlight has a parabolic cross section that is 6 ft wide at the opening and 2.5 ft deep at the vertex. How
far from the vertex is the focus? Round answer to two decimal places.
A) 0.90 ft B) 0.52 ft C) 0.21 ft D) 0.26 ft
Page 25
11.3 The Ellipse
1 Analyze Ellipses with Center at the Origin
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Match the graph to its equation.
1)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) x2
36 + y2
9 = 1B)
y2
9 – x2
36 = 1C)
x2
36 – y2
9 = 1D)
x2
9 + y2
36 = 1
2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) x2
4 + y2
16 = 1B)
y2
16 – x2
4 = 1C)
– y2
16 + x2
4 = 1D)
x2
16 + y2
4 = 1
Find the foci and vertices of the ellipse.
3) x2
25 + y2
4 = 1
A) foci at (– 21, 0) and ( 21, 0)
vertices at (–5, 0), (5, 0)
B) foci at (0, – 21) and (0, 21)
vertices at (0, –5), (0, 5)
C) foci at (–5
,
0) and (5
,
0)
vertices at (–25, 0), (25, 0)
D) foci at (0, –2) and (0, 2)
vertices at (0, –4), (0, 4)
Page 26
4) x2
16 + y2
49 = 1
A) foci at (0, – 33) and (0, 33)
vertices at (0, –7), (0, 7)
B) foci at (– 33, 0) and ( 33, 0)
vertices at (–7, 0), (7, 0)
C) foci at (0, –7) and (0, 7)
vertices at (0, –49), (0, 49)
D) foci at (0, 7) and (4
,
0)
vertices at (0, 49), (16, 0)
5) 25x2 + 36y2 = 900
A) foci at (– 11, 0) and ( 11, 0)
vertices at (–6, 0), (6, 0)
B) foci at (0, – 11) and (0, 11)
vertices at (0, –6), (0, 6)
C) foci at (–6
,
0) and (6
,
0)
vertices at (–36, 0), (36, 0)
D) foci at (0, –5) and (0, 5)
vertices at (0, –25), (0, 25)
6) 25x2 + 4y2 = 100
A) foci at (0, – 21) and (0, 21)
vertices at (0, –5), (0, 5)
B) foci at (– 21, 0) and ( 21, 0)
vertices at (–5, 0), (5, 0)
C) foci at (0, –5) and (0, 5)
vertices at (0, –25), (0, 25)
D) foci at (0, 5) and (2
,
0)
vertices at (0, 25) and (4, 0)
Find an equation for the ellipse.
7) Center at (0, 0); focus at (2
,
0); vertex at (3
,
0)
A) x2
9 + y2
5 = 1B)
x2
5 + y2
9 = 1C)
x2
4 + y2
5 = 1D)
x2
4 + y2
9 = 1
8) Center at (0, 0); focus at (–3
,
0); vertex at (7
,
0)
A) x2
49 + y2
40 = 1B)
x2
40 + y2
49 = 1C)
x2
9 + y2
40 = 1D)
x2
9 + y2
49 = 1
9) Center at (0, 0); focus at (–2
,
0); vertex at (7
,
0)
A) x2
49 + y2
45 = 1B)
x2
45 + y2
49 = 1C)
x2
4 + y2
45 = 1D)
x2
4 + y2
49 = 1
10) Vertices at (0, ±4); c = 2
A) x2
12 + y2
16 = 1B)
x2
16 + y2
12 = 1C)
x2
4 + y2
12 = 1D)
x2
4 + y2
16 = 1
11) Center at (0, 0); focus at (0, –6); vertex at (0, 7)
A) x2
13 + y2
49 = 1B)
x2
49 + y2
13 = 1C)
x2
36 + y2
13 = 1D)
x2
36 + y2
49 = 1
12) Foci at (0, ±5); a = 6
A) x2
11 + y2
36 = 1B)
x2
36 + y2
11 = 1C)
x2
25 + y2
11 = 1D)
x2
25 + y2
36 = 1
13) Focus at (–2
,
0); vertices at (±3
,
0)
A) x2
9 + y2
5 = 1B)
x2
5 + y2
9 = 1C)
x2
4 + y2
5 = 1D)
x2
4 + y2
9 = 1
Page 27
14) Focus at (0, –4); vertices at (0, ±7)
A) x2
33 + y2
49 = 1B)
x2
49 + y2
33 = 1C)
x2
16 + y2
33 = 1D)
x2
16 + y2
49 = 1
15) Foci at (±2
,
0); x–intercepts are ±4
A) x2
16 + y2
12 = 1B) x2
12 + y2
16 = 1C)
x2
4 + y2
12 = 1D)
x2
4 + y2
16 = 1
16) Foci at (0, ±4); y–intercepts are ±8
A) x2
48 + y2
64 = 1B) x2
64 + y2
48 = 1C)
x2
16 + y2
48 = 1D)
x2
16 + y2
64 = 1
17) Center (0, 0); major axis horizontal with length 20; length of minor axis is 4
A) x2
100 + y2
4 = 1B)
x2
4 + y2
100 = 1C)
x2
20 + y2
4 = 1D)
x2
400 + y2
16 = 1
18) Center (0, 0); major axis vertical with length 18; length of minor axis is 6
A) x2
9 + y2
81 = 1B)
x2
81 + y2
9 = 1C)
x2
6 + y2
81 = 1D)
x2
36 + y2
324 = 1
19) Foci at (0, ±3); length of the major axis is 14
A) x2
40 + y2
49 = 1B)
x2
49 + y2
40 = 1C)
x2
40 + y2
7 = 1D)
x2
49 + y2
7 = 1
Page 28
Graph the ellipse and locate the foci.
20) x2
16 + y2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) foci at (2 3, 0) and (–23, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) foci at (0, 2 3) and (0, –23)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) foci at ( 21, 0) and (–21, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) foci at (2 5, 0) and (–25, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 29
21) x2
4 + y2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) foci at (0, 2 3) and (0, –23)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) foci at (2 3, 0) and (–23, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) foci at (2 5, 0) and (–25, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) foci at ( 21, 0) and (–21, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 30
22) 4x2 + 16y2 = 64
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) foci at (2 3, 0) and (–23, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) foci at (0, 2 3) and (0, –23)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) foci at (2 5, 0) and (–25, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) foci at ( 21, 0) and (–21, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 31
23) 16x2 + 9y2 = 144
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) foci at (0, 7) and (0, –7)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) foci at ( 7, 0) and (–7, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) foci at (5
,
0) and (–5
,
0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) foci at (4
,
0) and (–4
,
0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 32
2 Analyze Ellipses with Center at (h, k)
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Write an equation for the graph.
1)
x
-5 5
y
5
-5
(-1, –2)
x
-5 5
y
5
-5
(-1, –2)
A) (x + 1)2
9 + (y + 2)2
4 = 1B)
(x + 2)2
9 + (y + 1)2
4 = 1
C) (x – 1)2
9 + (y – 2)2
4 = 1D)
(x + 1)2
4 + (y + 2)2
9 = 1
Find the center, foci, and vertices of the ellipse.
2) (x – 2)2
16 + (y + 1)2
9 = 1
A) center at (2
,
–1)
foci at (2 + 7
, –1), (2 – 7, –1)
vertices at (–2, –1), (6, –1)
B) center at (–1
,
2)
foci at (–1 + 7
, 2), (–1 – 7, 2)
vertices at (–2, –1), (6, –1)
C) center at (2
,
–1)
foci at (– 7
, –1), ( 7, –1)
vertices at (4, –1), (–4, –1)
D) center at (2
,
–1)
foci at (2 + 7
, 2), (2 – 7, 2)
vertices at (4, –1), (–4, –1)
3) 16(x – 1)2 + 9(y + 3)2 = 144
A) center at (1
,
–3)
foci at (1, –3 – 7
), (1, –3 + 7)
vertices at (1, 1), (1, –7)
B) center at (–3
,
1)
foci at (–3, 1 – 7
), (–3, 1 + 7)
vertices at (–3, 1), (–3, –7)
C) center at (–1
,
–3)
foci at (–1, –3 – 7
), (–1, –3 + 7)
vertices at (–1, 1), (–1, –7)
D) center at (2
,
–3)
foci at (2, –3 – 7
), (2, –3 + 7)
vertices at (2, 1), (2, –7)
Page 33
4) 3x2 + 4y2 – 42x + 24y + 171 = 0
A) (x – 7)2
4 + (y + 3)2
3 = 1
center: (7, –3); foci: (8, –3), (6, –3); vertices: (9, –3), (5, –3)
B) (x – 7)2
3 + (y + 3)2
4 = 1
center: (7, –3); foci: (8, –3), (6, –3); vertices: (9, –3), (5, –3)
C) (x – 7)2
4 + (y + 3)2
3 = 1
center: (–7, 3); foci: (–6, 3), (–8, 3); vertices: (–9, 3), (–5, 3)
D) (x – 7)2
3 + (y + 3)2
4 = 1
center: (–7, 3); foci: (–6, 3), (–8, 3); vertices: (–9, 3), (–5, 3)
5) 16x2 + y2 – 96x + 128 = 0
A) (x – 3)2 + y2
16 = 1
center: (3, 0); foci: (3, 15), (3, –15); vertices:(3, 4), (3, –4)
B) x2
16 + (y – 3)2 = 1
center: (3, 0); foci: (3, 15), (3, –15); vertices:(3, 4), (3, –4)
C) (x – 4)2 + y2
9 = 1
center: (4, 0); foci: (4, 2 2), (4, –22); vertices:(4, 3), (4, –3)
D) x2
9 + (y – 4)2 = 1
center: (4, 0); foci: (4, 2 2), (4, –15); vertices:(4, 3),
(4, –3)
Page 34
Graph the equation.
6) (x + 1)2
9 + (y – 2)2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 35
7) (x + 2)2
4 + (y – 1)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 36
8) 4(x – 1)2 + 9(y + 2)2 = 36
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 37
9) 16(x – 1)2 + 4(y + 2)2 = 64
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Find an equation for the ellipse described.
10) Center at (6
,
2); focus at (10
,
2); vertex at (12
,
2)
A) (x – 6)2
36 + (y – 2)2
20 = 1B)
(x + 6)2
36 + (y + 2)2
20 = 1
C) (x – 6)2
81 + (y + 2)2
8 = 2D)
(x + 6)2
16 – (y – 2)2
16 = 1
11) Vertices at (–2
,
4) and (14
,
4); focus at (12
,
4)
A) (x – 6)2
64 + (y – 4)2
28 = 1B)
(x – 4)2
49 + (y – 6)2
27 = 1
C) (x + 6)2
36 + (y + 4)2
28 = 1D)
(x – 6)2
100 – (y + 4)2
34 = 1
Page 38
Find an equation for the ellipse described. Graph the equation.
12) Foci at (–2
,
6) and (–2
,
0); length of major axis is 10
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) (y – 3)2
25 + (x + 2)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) (x – 3)2
25 + (y – 2)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) (y – 3)2
25 + (x – 2)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) (x – 3)2
16 + (y – 2)2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 39
13) Foci at (–4
,
–1) and (2
,
–1); length of major axis is 10
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) (x + 1)2
25 + (y + 1)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) (y – 2)2
25 + (x + 1)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) (x + 2)2
25 + (x – 1)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) (x – 2)2
25 + (y – 1)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 40
14) Vertices at (5, –4) and (5, 8); length of minor axis is 6
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) (x – 5)2
9 + (y – 2)2
36 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) (x – 5)2
36 + (y – 2)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) (x + 5)2
36 + (y + 2)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) (x – 5)2
9 – (y – 2)2
36 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 41
15) Foci at (0
,
3) and (–4
,
3); vertex at (–5
,
3)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) (x + 2)2
9 + (y – 3)2
5 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) (x + 2)2
5 + (y – 3)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) (x – 3)2
9 + (y + 2)2
5 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) (x – 3)2
5 + (y + 2)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 42
16) Center at (–2
,
4); focus at (–6
,
4); contains the point (–7
,
4)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) (x + 2)2
25 + (y – 4)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) (x + 2)2
9 + (y – 4)2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) (x + 4)2
25 + (y – 2)2
9 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) (x + 4)2
9 + (y – 2)2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 43
3 Solve Applied Problems Involving Ellipses
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Graph the function.
1) y = – 25
– 16x2
x
y
x
y
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Solve the problem.
2) A bridge is built in the shape of a semielliptical arch. It has a span of 118 feet. The height of the arch 26 feet
from the center is to be 5 feet. Find the height of the arch at its center.
A) 5.57 ft B) 5.13 ft C) 26.09 ft D) 11.35 ft
3) An arch for a bridge over a highway is in the form of a semiellipse. The top of the arch is 35 feet above
ground (the major axis). What should the span of the bridge be (the length of its minor axis) if the height
26 feet from the center is to be 10 feet above ground?
A) 54.26 ft B) 27.13 ft C) 182 ft D) 29.88 ft
Page 44
4) The orbit of a planet around a sun is an ellipse with the sun at one focus. The aphelion of a planet is its
greatest distance from the sun, its perihelion is its shortest distance, and its mean distance is the length of
the semimajor axis of the elliptical orbit. If a planet has a perihelion of 378.7 million miles and a mean
distance of 381 million miles, write an equation for the orbit of the planet around the sun.
A) x2
3812 + y2
380.9932 = 1B)
x2
381.0072 + y2
3812 = 1
C) x2
3812 + y2
2.32 = 1D)
x2
3812 + y2
378.72 = 1
5) An arch in the form of a semiellipse is 52 ft wide at the base and has a height of 20 ft. How wide is the arch
at a height of 12 ft above the base?
A) 41.6 ft B) 20.8 ft C) 35.5 ft D) 17.7 ft
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
6) A hall 130 feet in length was designed as a whispering gallery. If the ceiling is 25 feet high at the center,
how far from the center are the foci located?
7) A race track is in the shape of an ellipse 80 feet long and 60 feet wide. What is the width 32 feet from the
center?
Page 45
11.4 The Hyperbola
1 Analyze Hyperbolas with Center at the Origin
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Match the equation to the graph.
1) x2
4 – y2
9 = 1
A)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
B)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
C)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
D)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
Page 46
2) y2
4 – x2
16 = 1
A)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
B)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
C)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
D)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
Find an equation for the hyperbola described.
3) Vertices at (0, ±6); asymptotes at y = ± 3
8x
A) y2
36 – x2
256 = 1B)
y2
256 – x2
36 = 1C)
y2
36 – x2
64 = 1D)
y2
64 – x2
9 = 1
4) Vertices at (±3
,
0); foci at (±11
,
0)
A) x2
9 – y2
112 = 1B)
x2
112 – y2
9 = 1C)
x2
9 – y2
121 = 1D)
x2
121 – y2
9 = 1
Page 47
Find an equation for the hyperbola described. Graph the equation.
5) Center at (0, 0); focus at ( 61, 0); vertex at (6, 0)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) x2
36 – y2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) x2
25 – y2
36 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) y2
25 – x2
36 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) y2
36 – x2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 48
6) Center at (0, 0); vertex at (0, 5); focus at (0, 74)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) y2
25 – x2
49 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) y2
49 – x2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) x2
49 – y2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) x2
25 – y2
49 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 49
Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
7) x2
64 – y2
16 = 1
A) center at (0, 0)
transverse axis is x–axis
vertices at (–8, 0) and (8, 0)
foci at (– 45
, 0) and (4 5, 0)
asymptotes of y = – 1
2 and y = 1
2
B) center at (0, 0)
transverse axis is x–axis
vertices at (–4, 0) and (4, 0)
foci at (– 45
, 0) and (4 5, 0)
asymptotes of y = – 1
2 and y = 1
2
C) center at (0, 0)
transverse axis is y–axis
vertices at (0, –8) and (0, 8)
foci at (– 45, 0) and (4 5, 0)
asymptotes of y = – 1
2 and y = 1
2
D) center at (0, 0)
transverse axis is x–axis
vertices at (–8, 0) and (8, 0)
foci at (–4, 0) and (4, 0)
asymptotes of y = – 1
2 and y = 1
2
8) 100y2 – 49x2 = 4900
A) center at (0, 0)
transverse axis is y–axis
vertices at (0, –7) and (0, 7)
foci at (0, – 149) and (0, 149)
asymptotes of y = – 7
10 and y = 7
10
B) center at (0, 0)
transverse axis is x–axis
vertices: (–10, 0), (10, 0)
foci: (– 149, 0) , ( 149, 0)
asymptotes of y = – 7
10 and y = 7
10
C) center at (0, 0)
transverse axis is y–axis
vertices: (0, –7), (0, 7)
foci: (– 149, 0) , ( 149, 0)
asymptotes of y = – 7
10 and y = 7
10
D) center at (0, 0)
transverse axis is x–axis
vertices: (–7, 0), (7, 0)
foci: (–10, 0), (10, 0)
asymptotes of y = – 7
10 and y = 7
10
Write an equation for the hyperbola.
9)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
A) x2
4 – y2
16 = 1B)
y2
4 – x2
16 = 1C)
x2
16 – y2
4 = 1D)
y2
16 – x2
4 = 1
Page 50
10)
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
y
6
5
4
3
2
1
-1
-2
-3
-4
-5
-6
A) y2
9 – x2
4 = 1B)
x2
9 – y2
4 = 1C)
x2
4 – y2
9 = 1D)
y2
4 – x2
9 = 1
Page 51
Graph the hyperbola.
11) x2
9 – y2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 52
12) y2
9 – x2
25 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 53
13) 9x2 – 4y2 = 36
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 54
14) 25y2 – 9x2 = 225
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 55
15) 4x2 = 9y2 + 36
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 56
16) 16y2 = 4x2 + 64
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
2 Find the Asymptotes of a Hyperbola
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the asymptotes of the hyperbola.
1) x2
25 – y2
9 = 1
A) y = 3
5x and y = – 3
5xB)y
= 5
3x and y = – 5
3x
C) y = 9
25x and y = – 9
25xD)y
= 25
9x and y = – 25
9x
Page 57
2) y2 – x2 = 4
A) y = x and y = – xB)y =2x and y = – 2x
C) y = 1
2x and y = – 1
2xD)y
= 1
4x and y = – 1
4x
3) (x + 2)2
4 – (y + 1)2
25 = 1
A) y + 1 = 5
2(x + 2) and y + 1 = – 5
2(x + 2) B) y + 1 = 2
5(x + 2) and y + 1 = – 2
5(x + 2)
C) y = 5
2(x + 2) and y = – 5
2(x + 2) D) y + 2 = 5
2(x + 1) and y + 2 = – 5
2(x + 1)
4) x2 – y2 – 6x + 4y – 20 = 0
A) y – 2 = (x – 3) and y – 2 = – (x – 3) B) y – 2 = 1
5(x – 3) and y – 2 = – 1
5(x – 3)
C) y + 3 = (x + 2) and y + 3 = – (x + 2) D) y –3=(x –2) and y – 3 = – (x –2)
3 Analyze Hyperbolas with Center at (h, k)
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find an equation for the hyperbola described.
1) Vertices (1
2, –3) and (– 9
2, –3); asymptotes y + 3 = ± 6
5(x + 2)
A) 4(x + 2)2
25 – (y + 3)2
9 = 1B)
(x + 2)2
9 – 4(y + 3)2
25 = 1
C) (y + 3)2
9 – 4(x + 2)2
25 = 1D)
4(x – 2)2
25 – (y – 3)2
9 = 1
2) center at (7
,
4); focus at (0
,
4); vertex at (6
,
4)
A) (x – 7)2 – (y – 4)2
48 = 1B)
(x – 7)2
48 – (y – 4)2 = 1
C) (x – 4)2 – (y – 7)2
48 = 1D)
(x – 4)2
48 – (y – 7)2 = 1
3) Vertices at (0, ±4); asymptotes at y = ± 2
5x
A) y2
16 – x2
100 = 1B)
y2
100 – x2
16 = 1C) y2
16 – x2
25 = 1D)
y2
25 – x2
4 = 1
4) Vertices at (±4
,
0); foci at (±8
,
0)
A) x2
16 – y2
48 = 1B)
x2
48 – y2
16 = 1C)
x2
16 – y2
64 = 1D)
x2
64 – y2
16 = 1
Page 58
Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
5) (x + 4)2
9 – (y – 1)2
4 = 1
A) center at (–4
,
1)
transverse axis is parallel to x–axis
vertices at (–7, 1) and (–1, 1)
foci at (–4 – 13, 1) and (–4 + 13, 1)
asymptotes of y – 1 = – 2
3(x + 4) and y – 1 = 2
3(x + 4)
B) center at (1
,
–4)
transverse axis is parallel to x–axis
vertices at (–2, –4) and (4, –4)
foci at (1 – 13, –4) and (1 + 13, –4)
asymptotes of y + 4 = – 2
3(x – 1) and y + 4 = 2
3(x – 1)
C) center at (–4
,
1)
transverse axis is parallel to y–axis
vertices at (–4, –2) and (–4, 4)
foci at (–4, 1 – 13) and (–4, 1 + 13)
asymptotes of y + 1 = – 3
2(x – 4) and y + 1 = 3
2(x – 4)
D) center at (–4
,
1)
transverse axis is parallel to x–axis
vertices at (–6, 1) and (–2, 1)
foci at (–4 – 13, 1) and (–4 + 13, 1)
asymptotes of y – 1 = – 3
2(x + 4) and y – 1 = 3
2(x + 4)
Page 59
6) (x – 4)2 – 25(y – 3)2 = 25
A) center at (4
,
3)
transverse axis is parallel to x–axis
vertices at (–1, 3) and (9, 3)
foci at (4 – 26, 3) and (4 + 26, 3)
asymptotes of y – 3 = – 1
5(x – 4) and y – 3 = 1
5(x – 4)
B) center at (3
,
4)
transverse axis is parallel to x–axis
vertices at (–2, 4) and (8, 4)
foci at (3 – 26, 4) and (3 + 26, 4)
asymptotes of y – 4 = – 1
5(x – 3) and y – 4 = 1
5(x – 3)
C) center at (4
,
3)
transverse axis is parallel to y–axis
vertices at (4, –2) and (4, 8),
foci at (4, 3 – 26) and (4, 3 + 26),
asymptotes of y + 3 = – 5(x + 4) and y + 3 = 5(x + 4)
D) center at (4
,
3)
transverse axis is parallel to x–axis
vertices at (3, 3) and (5, 3)
foci at (4 – 26, 3) and (4 + 26, 3)
asymptotes of y – 3 = – 5(x – 4) and y – 3 = 5(x – 4)
7) x2 – 4y2 + 8x – 8y + 8 = 0
A) center at (–4
,
–1)
transverse axis is parallel to x–axis
vertices at (–6, –1) and (–2, –1)
foci at (–4 – 5, –1) and (–4 + 5, –1)
asymptotes of y + 1 = – 1
2(x + 4) and y + 1 = 1
2(x + 4)
B) center at (–1
,
–4)
transverse axis is parallel to x–axis
vertices at (–3, –4) and (1, –4)
foci at (–1 – 5
, –4) and (–1 + 5, –4)
asymptotes of y + 4 = – 1
2(x + 1) and y + 4 = 1
2(x + 1)
C) center at (–4
,
–1)
transverse axis is parallel to y–axis
vertices at (–4, –3) and (–4, 1)
foci at (–4, –1 – 5) and (–4, –1 + 5)
asymptotes of y – 1 = – 2(x – 4) and y – 1 = 2(x – 4)
D) center at (–4
,
–1)
transverse axis is parallel to x–axis
vertices at (–5, –1) and (–3, –1)
foci at (–4 – 5
, –1) and (–4 + 5, –1)
asymptotes of y + 1 = – 2(x + 4) and y + 1 = 2(x + 4)
Page 60
Graph the hyperbola.
8) (x – 2)2
4 – (y + 1)2
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 61
9) (y + 2)2
9 – (x – 2)2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 62
10) (x – 1)2 – 4(y – 1)2 = 4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 63
11) (y – 1)2 – 9(x + 4)2 = 9
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 64
4 Solve Applied Problems Involving Hyperbolas
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Graph the function.
1) y = –3 + x2
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Solve the problem.
2) Two recording devices are set 2800 feet apart, with the device at point A to the west of the device at poin
t
B. At a point on a line between the devices, 200 feet from point B, a small amount of explosive is
detonated. The recording devices record the time the sound reaches each one. How far directly north of
site B should a second explosion be done so that the measured time difference recorded by the devices is
the same as that for the first detonation?
A) 433.33 ft B) 4996 ft C) 1385.64 ft D) 1108.05 ft
Page 65
3) The roof of a building is in the shape of the hyperbola y2 – x2 = 50, where x and y are in meters. Refer to
the figure and determine the height h of the outside walls.
a = b = 7 m
A) 9.9 m B) 99 m C) –1 m D) 43 m
4) The roof of a building is in the shape of the hyperbola y2 – x2 = 56, where x and y are in meters.
Determine the distance, w, the outside walls are apart, if the height of each wall is 10 m.
A) 13.3 m B) 44 m C) 6.65 m D) 12.5 m
5) A comet follows the hyperbolic path described by x2
25 – y2
17 = 1, where x and y are in millions. If the sun is
the focus of the path, how close to the sun is the vertex of the path?
A) 1.5 million B) 6.5 million C) 5 million D) 42 million
Page 66
6) A satellite following the hyperbolic path shown in the picture turns rapidly at (0, 4) and then moves closer
and closer to the line y = 8
5x as it gets farther from the tracking station at the origin. Find the equation that
describes the path of the rocket if the center of the hyperbola is at (0, 0).
(0, 4)
y = 8
5x
A) y2
16 – x2
25
4
= 1B)
x2
16 – y2
16
5
2 = 1C)
y2
25
4
– x2
16 = 1D)
x2
16
5
2 – y2
16 = 1
11.5 Rotation of Axes; General Form of a Conic
1 Identify a Conic
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Identify the equation without completing the square.
1) 4x2 – 4x + y – 1 = 0
A) parabola B) ellipse C) hyperbola D) not a conic
2) 2y2 – 2x – 2y = 0
A) parabola B) ellipse C) hyperbola D) not a conic
3) 2x2 + 3y2 – 4x + 2y = 0
A) ellipse B) parabola C) hyperbola D) not a conic
4) 3x2 + 2y2 + 6x + 3 = 0
A) ellipse B) parabola C) hyperbola D) not a conic
5) 3x2 – 3y2 + 5x + 2y + 2 = 0
A) hyperbola B) parabola C) ellipse D) not a conic
6) y2 – 3x2 – 2x + 2y + 2 = 0
A) hyperbola B) parabola C) ellipse D) not a conic
Page 67
2 Use a Rotation of Axes to Transform Equations
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Determine the appropriate rotation formulas to use so that the new equation contains no xy–term.
1) x2 + 2xy + y2 – 8 = 0
A) x = 2
2(x’ – y’) and y = 2
2(x’ + y’)
B) x = –y’ and y = x’
C) x = 2 + 2
2x’ – 2 – 2
2y’ and y = 2 – 2
2x’ + 2 + 2
2y’
D) x = 1
2x’ – 3
2y’ and y = 3
2x’ + 1
2y’
2) 2x2 + 4xy + 2y2 – 8x + 8y = 0
A) x = 2
2(x’ – y’) and y = 2
2(x’ + y’)
B) x = –y’ and y = x’
C) x = 2 + 2
2x’ – 2 – 2
2y’ and y = 2 – 2
2x’ + 2 + 2
2y’
D) x = 1
2x’ – 3
2y’ and y = 3
2x’ + 1
2y’
3) 9x2 – 4xy + 5y2 – 8x + 8y = 0
A) x = 2 – 2
2x’ – 2 + 2
2y’ and y = 2 + 2
2x’ + 2 – 2
2y’
B) x = 2
2(x’ – y’) and y = 2
2(x’ + y’)
C) x = –y’ and y = x’
D) x = 1
2x’ – 3
2y’ and y = 3
2x’ + 1
2y’
4) 14x2 + 12 3xy + 2y2 – 24 = 0
A) x = 1
2(3x’ – y’) and y = 1
2(x’ + 3y’) B) x = 1
2(3x’ + y’) and y = 1
2(x’ – 3y’)
C) x = 1
2(x’ – 3y’) and y = 1
2(3x’ + y’) D) x = 1
2(2x’ – y’) and y = 1
2(x’ + 2y’)
5) 15
4x2 + 11xy + 12y2 + 45y + 23 = 0
A) x = 5
5(x’ – 2y’) and y = 5
5(2x’ + y’) B) x = 5
5(2x’ + y’) and y = 5
5(x’ – 2y’)
C) x = 5
5(x’ + 2y’) and y = 5
5(2x’ – y’) D) x = 5
5(2x’ – y’) and y = 5
5(x’ + 2y’)
Page 68
3 Analyze an Equation Using a Rotation of Axes
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Rotate the axes so that the new equation contains no xy–term. Discuss the new equation.
1) 24xy – 7y2 + 36 = 0
A) θ = 36.9°
y’2
4 – 4x‘2
9 = 1
hyperbola
center at (0, 0)
transverse axis is the y’–axis
vertices at (0, ±2)
B) θ=53.1°
y’2
4 – 4x‘2
9 = 1
hyperbola
center at (0, 0)
transverse axis is the y’–axis
vertices at (0, ±2)
C) θ = 36.9°
y’2
9 – x’2
16 = 1
hyperbola
center at (0, 0)
transverse axis is the y’–axis
vertices at (0, ±3)
D) θ=36.9°
4y‘2
9 – x’2
4 = 1
hyperbola
center at (0, 0)
transverse axis is the y’–axis
vertices at (0, ± 3
2)
2) x2 + 2xy + y2 – 8x + 8y = 0
A) θ = 45°
x’2 = –42
y’
parabola
vertex at (0, 0)
focus at (0, –2)
B) θ=45°
y’2 = –42x’
parabola
vertex at (0, 0)
focus at (–2, 0)
C) θ = 36.9°
x’2
4 + y’2
4 = 1
ellipse
center (0, 0)
major axis is x’–axis
vertices at (±2, 0)
D) θ=36.9°
x’2
4 + y’2
2 = 1
ellipse
center (0, 0)
major axis is x’–axis
vertices at (±2, 0)
Page 69
3) 31x2 + 10 3xy + 21y2 –144 = 0
A) θ = 30°
x’2
4 + y’2
9 = 1
ellipse
center at (0, 0)
major axis is y’–axis
vertices at (0, ±3)
B) θ=45°
y’2 = –42
x’
parabola
vertex at (0, 0)
focus at (–2, 0)
C) θ = 45°
x’2 = –42
y’
parabola
vertex at (0, 0)
focus at (0, –2)
D) θ=36.9°
x’2
9 + y’2
4 = 1
ellipse
center at (0, 0)
major axis is x’–axis
vertices at (±3, 0)
4) xy +16 = 0
A) θ = 45°
y’2
32 – x’2
32 = 1
hyperbola
center at (0, 0)
transverse axis is y’–axis
vertices at (0, ±42
)
B) θ=45°
y’2 = –32x’
parabola
vertex at (0, 0)
focus at (–8, 0)
C) θ = 45°
y’2
32 + x’2
32 = 1
ellipse
center at (0, 0)
major axis is y’–axis
vertices at (0, ±42
)
D) θ=36.9°
x’2
4 + y’2
2 = 1
ellipse
center at (0, 0)
major axis is the x’–axis
vertices at (±2, 0)
Page 70
5) x2 + xy + y2 – 3y – 6 = 0
A) θ = 45°
x’ – 2
2
2
5 +
y’ – 32
2
2
15 = 1
ellipse
center at 2
2, 32
2
major axis is y’–axis
vertices at 2
2, – 32
2 and 2
2, 92
2
B) θ=45°
y’2 = –18x’
parabola
vertex at (0, 0)
focus at – 9
2, 0
C) θ = 45°
x’2
6 – y’2
8 = 1
hyperbola
center at (0, 0)
transverse axis is the x’–axis
vertices at (±6, 0)
D) θ=45°
x’2
3 + y’2
4 = 1
ellipse
center at (0, 0)
major axis is y’–axis
vertices at (0, ±2)
Page 71
6) 17x2 – 12xy + 8y2 – 68x + 24y –12 = 0
A) θ = 63.4°
x’ – 25
5
2
16 +
y’ + 45
5
2
4 = 1
ellipse
center at (25
5, – 45
5)
major axis is x’–axis
vertices at (4 + 25
5, – 45
5) and (–4 + 25
5, – 45
5)
B) θ = 63.4°
x’2 = –16y’
parabola
vertex at (0, 0)
focus at (0, –4)
C) θ = 63.4°
x’2
16 – y’2
4 = 1
hyperbola
center at (0, 0)
transverse axis is the x’–axis
vertices at (±4, 0)
D) θ = 26.6°
x’2
4 + y’2
16 = 1
ellipse
center at (0, 0)
major axis is y’–axis
vertices at (0, ±4)
7) 5x2 – 6xy + 5y2 – 8 = 0
A) θ = 45°
x’2
4 + y’2 = 1
ellipse
center at (0, 0)
major axis is the x’–axis
vertices at (±2, 0)
B) θ=45°
y’2 = –4x’
parabola
vertex at (0, 0)
focus at (–1, 0)
C) θ = 45°
x’2 = –4y’
parabola
vertex at (0, 0)
focus at (0, –1)
D) θ=45°
x’2
4 – y’2 = 1
hyperbola
center at (0, 0)
transverse axis is the x’–axis
vertices at (±2, 0)
Page 72
Rotate the axes so that the new equation contains no xy–term. Graph the new equation.
8) 24xy – 7y2 + 36 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 73
9) x2 + 2xy + y2 – 8x + 8y = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 74
10) 31x2 + 10 3xy + 21y2 –144 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 75
11) xy +16 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 76
12) x2 + xy + y2 – 3y – 6 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 77
13) 17x2 – 12xy + 8y2 – 68x + 24y –12 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
Page 78
14) 5x2 – 6xy + 5y2 – 8 = 0
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
A)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
B)
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8
y
10
8
6
4
2
-2
-4
-6
-8
-10
4 Identify Conics without a Rotation of Axes
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Identify the equation without applying a rotation of axes.
1) x2 + 12xy + 36y2 – 4x – 2y – 6 = 0
A) parabola B) ellipse C) hyperbola D) not a conic
2) 2x2 + 12xy + 36y2 – 4x + 4y + 3 = 0
A) ellipse B) parabola C) hyperbola D) not a conic
3) 2x2 – 11xy + 2y2 – 4x + 3y – 5 = 0
A) hyperbola B) ellipse C) parabola D) not a conic
4) x2 + 4xy – 4y2 + 2x – 4y – 6 = 0
A) hyperbola B) ellipse C) parabola D) not a conic
5) 3x2 + 3xy + 3y2 + 3x + 3y + 10 = 0
A) ellipse B) hyperbola C) parabola D) not a conic
6) 7x2 + 7xy + 2y2 – 3x – 2y + 10 = 0
A) ellipse B) hyperbola C) parabola D) not a conic
Page 79
11.6 Polar Equations of Conics
1 Analyze and Graph Polar Equations of Conics
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Identify the conic that the polar equation represents. Also, give the position of the directrix.
1) r = 9
1 – 3 cos θ
A) hyperbola, directrix perpendicular to the polar axis 3 left of the pole
B) hyperbola, directrix perpendicular to the polar axis 3 right of the pole
C) ellipse, directrix perpendicular to the polar axis 3 left of the pole
D) ellipse, directrix perpendicular to the polar axis 3 right of the pole
2) r = 6
2 + 2 sin θ
A) parabola, directrix parallel to the polar axis 3 above the pole
B) parabola, directrix perpendicular to the polar axis 3 right of the pole
C) hyperbola, directrix parallel to the polar axis 3 above the pole
D) hyperbola, directrix perpendicular to the polar axis 3 right of the pole
3) r = 6
6 – 3 sin θ
A) ellipse, directrix parallel to the polar axis 2 below the pole
B) ellipse, directrix perpendicular to the polar axis 2 left of the pole
C) ellipse, directrix perpendicular to the polar axis 2 right of the pole
D) ellipse, directrix parallel to the polar axis 2 above the pole
4) r = 6
3 – 4 cos θ
A) hyperbola, directrix is perpendicular to the polar axis at a distance 3
2 units to the left of the pole
B) ellipse, directrix is perpendicular to the polar axis at a distance 3
2 units to the right of the pole
C) ellipse, directrix is perpendicular to the polar axis at a distance 3 units to the left of the pole
D) hyperbola, directrix is perpendicular to the polar axis at a distance 3 units to the right of the pole
Discuss the equation and graph it.
5) r = 4
2 – 2 cos θ
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
Page 80
A) parabola; directrix perpendicular to
the polar axis 2 units to left of pole
focus (0, 0), vertex 1, π
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
B) parabola; directrix perpendicular to
polar axis 2 units to right of pole
focus (0, 0), vertex 1, 0
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
C) parabola; directrix parallel to
the polar axis 2 units above pole
focus (0, 0), vertex 1, π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
D) parabola; directrix parallel to
the polar axis 2 units below pole
focus (0, 0), vertex 1, 3π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
6) r = 4
2 – sin θ
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
Page 81
A) ellipse; directrix parallel to
the polar axis 4 units below pole
center 4
3, π
2
vertices 4, π
2, 4
3, 3π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
B) ellipse; directrix perpendicular to
polar axis 4 units right of pole
center – 4
3, 0
vertices 4, π, 4
3, 0
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
C) ellipse; directrix perpendicular to
the polar axis 4 units left of pole
center 4
3, 0
vertices 4
3, π, 4, 0
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
D) ellipse; directrix parallel to
the polar axis 4 units above pole
center – 4
3, π
2
vertices – 4
3, 3π
2, 4, 3π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
7) r = 3
2 + 4 sin θ
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
Page 82
A) hyperbola; directrix parallel to
the polar axis 3
4 unit above the pole
vertices 1
2, π
2, – 3
2, 3π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
B) hyperbola, directrix perpendicular to
the polar axis 3
4 unit right of the pole
vertices 1
2, 0 , – 3
2, π
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
C) ellipse, directrix perpendicular to
the polar axis 3
2 unit left of the pole
vertices 3
2, 0 , 1
2, π
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
D) ellipse, directrix parallel to
the polar axis 3
2 unit below the pole
vertices 3
2, π
2, 1
2, 3π
2
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
r
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
Find a polar equation for the conic. A focus is at the pole.
8) e = 1; directrix is parallel to the polar axis 1 above the pole
A) r = 1
1 + sin θB) r = 1
1 – sin θC) r = 1
1 + cos θD) r = 1
1 – cos θ
9) e = 3
5; directrix is perpendicular to the polar axis 2 to the left of the pole
A) r = 6
5 – 3 cos θB) r = 6
5 + 3 cos θC) r = 10
5 – 3 cos θD) r = 10
5 + 3 cos θ
10) e = 3; directrix is perpendicular to the polar axis 2 to the right of the pole
A) r = 6
1 + 3 cos θB) r = 6
1 – 3 cos θC) r = 6
1 + 3 sin θD) r = 6
1 – 3 sin θ
Page 83
Solve the problem.
11) A planet travels around the sun in an elliptical orbit given approximately by r = (2.477) 107
1 – 0.393 cos θ, where r is
measured in miles and the Sun is at the pole. Find the distance from the planet to the Sun at aphelion
(greatest distance from the Sun.)
A) 4.08 × 107mi B) 1.78 × 107mi C) 4.08 × 106mi D) 2.48 × 107mi
12) A planet travels around the sun in an elliptical orbit given approximately by r = (2.398) 107
1 – 0.235 cos θ, where r is
measured in miles and the Sun is at the pole. Find the distance from the planet to the Sun at perihelion
(shortest distance from the Sun.)
A) 1.94 × 107mi B) 3.13 × 107mi C) 1.94 × 106mi D) 2.40 × 107mi
2 Convert the Polar Equation of a Conic to a Rectangular Equation
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Convert the polar equation to a rectangular equation.
1) r = 8
4 – 4 cos θ
A) y2 = 4x + 4B)y
2 = –4x + 4C)x
2 = 4y + 4D)x
2 = –4y + 4
2) r = 6
2 – 2 sinθ
A) x2 = 6y + 9B)y
2 = 6x + 9C)y
2 = –6x + 9D)x
2 = –6y + 9
3) r = 4
4 + cos θ
A) 15x2 + 16y2 + 8x – 16 = 0 B) 17x2 + 16y2 – 8x – 16 = 0
C) 16x2 + 15y2 + 8y – 16 = 0 D) 16x2 + 16y2 + 8x – 16 = 0
Page 84
4) r = 12
4 + sin θ
A) 16x2 + 15y2 + 24y – 144 = 0 B) 15x2 + 16y2 + 24x – 144 = 0
C) 17x2 + 16y2 – 24x – 144 = 0 D) 16x2 + 16y2 + 24x – 144 = 0
5) r(3 + cos θ)= 6
A) 8x2 + 9y2 + 12x – 36 = 0 B) 10x2 + 9y2 – 12x – 36 = 0
C) 9x2 + 8y2 + 12y – 36 = 0D)9x
2 + 9y2 + 12x – 36 = 0
6) r = 6 sec θ
3 sec θ + 1
A) 8x2 + 9y2 + 12x – 36 = 0 B) 10x2 + 9y2 – 12x – 36 = 0
C) 9x2 + 8y2 + 12y – 36 = 0D)9x
2 + 9y2 + 12x – 36 = 0
7) r = 4 sec θ
sec θ + 2
A) 3x2 – y2 –16x + 16 = 0B)3x
2 – y2 + 16 = 0
C) 3y2 – x2 –16y + 16 = 0D)3y
2 – x2 + 16 = 0
Page 85
11.7 Plane Curves and Parametric Equations
1 Graph Parametric Equations by Hand
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Graph the curve whose parametric equations are given.
1) x = 3t, y = t + 2; –2 ≤ t ≤ 3
x
-10 10
y
10
-10
x
-10 10
y
10
-10
A)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
B)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
C)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
D)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
Page 86
2) x = 2t – 1, y = t2 + 5; –4 ≤ t ≤ 4
x
-10 10
y
10
-10
x
-10 10
y
10
-10
A)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
B)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
C)
x
-10 10
y
50
-50
x
-10 10
y
50
-50
D)
x
-10 10
y
50
-50
x
-10 10
y
50
-50
Page 87
3) x = t3 + 1, y = t3 – 15; –2 ≤ t ≤ 2
x
-30 30
y
40
-40
x
-30 30
y
40
-40
A)
x
-30 30
y
40
-40
x
-30 30
y
40
-40
B)
x
-30 30
y
40
-40
x
-30 30
y
40
-40
C)
x
-30 30
y
40
-40
x
-30 30
y
40
-40
D)
x
-30 30
y
40
-40
x
-30 30
y
40
-40
Page 88
4) x = 6 sin t, y = 6 cos t; 0 ≤ t ≤ 2π
x
-10 10
y
10
-10
x
-10 10
y
10
-10
A)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
B)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
C)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
D)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
Page 89
5) x = 3 tan t, y = 4 sec t; 0 ≤ t ≤ 2π
x
-10 10
y
10
-10
x
-10 10
y
10
-10
A)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
B)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
C)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
D)
x
-10 10
y
10
-10
x
-10 10
y
10
-10
Page 90
6) x = t2, y = t + 6; 0 ≤ t ≤ 4
x
-20 20
y
10
-10
x
-20 20
y
10
-10
A)
x
-20 20
y
10
-10
x
-20 20
y
10
-10
B)
x
-20 20
y
10
-10
x
-20 20
y
10
-10
C)
x
-20 20
y
10
-10
x
-20 20
y
10
-10
D)
x
-20 20
y
10
-10
x
-20 20
y
10
-10
7) x = –sec t, y = tan t; – π
2 < t < π
2
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 91
A) B)
C) D)
8) x = 3 cos t, y = – 3 sin t; π
2 ≤ t ≤ 3π
2
x
–5–4–3–2–1 12345
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
–5–4–3–2–1 12345
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 92
A)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
B)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
C)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
D)
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
x
-5 -4 -3 -2 -1 1 2 3 4 5
y
5
4
3
2
1
-1
-2
-3
-4
-5
Page 93
9) x = t, y = 4t + 2; 0 ≤ t ≤ 4
x
-5 5
y
20
-20
x
-5 5
y
20
-20
A)
x
-5 5
y
20
-20
x
-5 5
y
20
-20
B)
x
-5 5
y
20
-20
x
-5 5
y
20
-20
C)
x
-5 5
y
20
-20
x
-5 5
y
20
-20
D)
x
-5 5
y
20
-20
x
-5 5
y
20
-20
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
The parametric equations of four curves are given. Graph each of them, indicating the orientation.
10) C1: x = 7sin t, y = 7 – 7cos2 t; π
2 ≤ t ≤ 3π
2
C2: x = ln t, y = ln t2; e–4 ≤ t ≤ e3
C3: x = t2 – 8, y = t – 3; –4 ≤ t ≤ 4
C4: x = t – 5, y = t + 2; –4 ≤ t ≤ 7
Page 94
2 Graph Parametric Equations Using a Graphing Utility
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Use a graphing utility to graph the curve defined by the given parametric equations.
1) x = t + 2, y = 3t – 1; 0 ≤t ≤ 3
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
A) B)
C) D)
Page 95
2) x = 2t2, y = t + 2; –∞ < t < ∞
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
A) B)
C) D)
Page 96
3) x = 3 cos t, y = 2 sin t; 0 ≤ t ≤ 2π
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
A) B)
C) D)
3 Find a Rectangular Equation for a Curve Defined Parametrically
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find a rectangular equation for the plane curve defined by the parametric equations.
1) x = 2t, y = t + 1; –2 ≤ t ≤ 3
A) y = 1
2x + 1; for x in –4 ≤ x ≤ 6B)y = –2x +1; for x in –∞
<
x
<
∞
C) y = 1
2x – 1; for x in –∞ < x < ∞D) y = x2 + 1; for x in –2 ≤ x ≤ 2
Page 97
2) x = 2t – 1, y = t2 + 6; –4 ≤ t ≤ 4
A) y = 1
4x2 + 1
2x + 25
4; for x in –9 ≤ x ≤ 7B)y = 1
2x2 + 1; for x in –6 ≤ x ≤ 4
C) y = – 1
2x + 30; for x in –6 ≤ x ≤ 4D)y = x2 + 1; for x in –2 ≤ x ≤ 2
3) x = t + 4, y = t2; –∞ < t < ∞
A) y = x2 – 8x + 16 B) y = x2 + 16 C) y = x
– 4 D) y = x + x + 4
4) x = t
, y = 2t + 5; 0 < t < ∞
A) y = 2x2 + 5; for x in 0 < x < ∞B) y = 2x
+ 5; for x in 0 < x < ∞
C) y = 2x2 – 5; for x in 0 < x < ∞D) y = 2x
– 5; for x in 0 < x < ∞
5) x = t3 + 1, y = t3 – 10; –2 ≤ t ≤ 2
A) y = x – 11; for x in –7 ≤ x ≤ 9B)y = – x –11; for x in –7 ≤ x ≤ 9
C) y = – x2; for x in –4 ≤ x ≤ 4D)y = x3; for x in –3 ≤ x ≤ 1
6) x = et, y = e2t; –∞ < t < ∞
A) y = x2B) y = 2xC) y =ln x D) x =ln y
7) x = 8 sin t, y = 8 cos t; 0 ≤ t ≤ 2π
A) x2 + y2 = 64; for x in –8 ≤ x ≤ 8B)y
2 – x2 = 64; for x in –∞ < x < ∞
C) y = a
2 – x2 = 64; for x in –∞ < x < ∞D) y = x2 – 9; for x in –2 ≤ x ≤ 2
8) x = 5 tan t, y = 4 sec t; 0 ≤ t ≤ 2π
A) y2
16 – x2
25 = 1; for x in –∞ < x < ∞B) y2
16 + x2
25 = 1; for x in –∞ < x < ∞
C) y = 41 + x2
25 ; for x in –∞ < x < ∞D) y = x2 – 9; for x in –3 ≤ x ≤ 3
9) x = 5 cos t, y = –2 sin t; 0 ≤ t ≤ 2π
A) 4x2 + 25y2 = 100; –5 ≤ x ≤ 5B)4x
2 + 25y2 = 1; – 1
5 ≤ x ≤ 1
5
C) 4x2 – 25y2 = 100; x ≥ 5D)4x
2 – 25y2 = 1; x ≥ 1
2
Solve the problem.
10) The position of a projectile fired with an initial velocity v0feet per second and at an angle θ to the
horizontal at the end of t seconds is given by the parametric equations x = (v0 cos θ)t,
y = (v0 sin θ)t – 16t2. Suppose the initial velocity is 12 feet per second. Obtain the rectangular equation of
the trajectory and identify the curve.
A) y = – 1
9 x2
cos2 θ
+ (tan θ)x; parabola B) y = – 1
9 x2
cos2 θ
+ (tan θ)x; ellipse
C) y = 1
9 x2
cos2 θ
+ (cot θ)x; parabola D) y = – 1
9 x2
cos2 θ
+ (cot θ)x; hyperbola
Page 98
4 Use Time as a Parameter in Parametric Equations
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
1) Ron throws a ball straight up with an initial speed of 40 feet per second from a height of 3 feet. Find
parametric equations that describe the motion of the ball as a function of time. How long is the ball in the
air? When is the ball at its maximum height? What is the maximum height of the ball?
A) x = 0, y = –16t2 + 40t + 3
2.573 sec, 1.25 sec,
28 feet
B) x = 0, y = –16t2 + 40t + 3
5.146 sec, 1.25 sec,
25 feet
C) x = 0, y = –16t2 + 40t + 3
2.423 sec, 1.25 sec,
2.985 feet
D) x = 0, y = –16t2 + 40t + 3
4.845 sec, 1.25 sec,
190.855 feet
2) A baseball pitcher throws a baseball with an initial speed of 131 feet per second at an angle of 20° to the
horizontal. The ball leaves the pitcher’s hand at a height of 4 feet. Find parametric equations that describe
the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its
maximum height? What is the maximum height of the ball?
A) x = 123.1t, y = –16t2 + 44.8t + 4
2.887 sec, 1.4 sec,
35.36 feet
B) x = 123.1t, y = –16t2 + 44.8t + 4
5.773 sec, 1.4 sec,
31.36 feet
C) x = 123.1t, y = –16t2 + 44.8t + 4
2.708 sec, 1.4 sec,
3.986 feet
D) x = 123.1t, y = –16t2 + 44.8t + 4
5.415 sec, 1.4 sec,
254.88 feet
3) A baseball player hit a baseball with an initial speed of 170 feet per second at an angle of 40° to the
horizontal. The ball was hit at a height of 5 feet off the ground. Find parametric equations that describe the
motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum
height? What is the distance the ball traveled?
A) x = 130.22t, y = –16t2 + 109.31t + 5
6.877 sec, 3.416 sec,
895.523 feet
B) x = 130.22t, y = –16t2 + 109.31t + 5
13.755 sec, 3.416 sec,
1791.176 feet
C) x = 130.22t, y = –16t2 + 109.31t + 5
6.786 sec, 3.416 sec,
883.673 feet
D) x = 130.22t, y = –16t2 + 109.31t + 5
6.877 sec, 3.416 sec,
1508.415 feet
4) Rachel’s bus leaves at 2:15 PM and accelerates at the rate of 4 meters per second per second. Rachel, who
can run 5 meters per second, arrives at the bus station 5 seconds after the bus has left. Find parametric
equations that describe the motions of the bus and Rachel as a function of time. Determine algebraically
whether Rachel will catch the bus. If so, when?
A) Bus: x1 = 2t2, y1 = 1; Rachel: x2 = 5(t – 5), y2 = 3
Rachel won’t catch the bus.
B) Bus: x1 = 2t2, y1 = 1; Rachel: x2 = 5(t + 5), y2 = 3
Rachel won’t catch the bus.
C) Bus: x1 = 2t2, y1 = 1; Rachel: x2 = 5(t – 5), y2 = 3
Rachel will catch the bus at 2:20 PM
D) Bus: x1 = 4t2, y1 = 1; Rachel: x2 = 5
2(t – 5), y2 = 3
Rachel will catch the bus at 2:19 PM
Page 99
5) Rachel’s bus leaves at 5:35 PM and accelerates at the rate of 3 meters per second per second. Rachel, who
can run 7 meters per second, arrives at the bus station 4 seconds after the bus has left. Find parametric
equations that describe the motions of the bus and Rachel as a function of time, and simulate the motion of
the bus and Rachel by simultaneously graphing these equations.
x
100
y
5
4
3
2
1
x
100
y
5
4
3
2
1
A) Bus: x1 = 3
2t2, y1 = 2;
Rachel: x2 = 7(t – 4), y2 = 4
x
100
y
5
4
3
2
1
x
100
y
5
4
3
2
1
B) Bus: x1 = 3
2t2, y1 = 2;
Rachel: x2 = 7(t + 4), y2 = 4
x
100
y
5
4
3
2
1
x
100
y
5
4
3
2
1
C) Bus: x1 = 3t2, y1 = 2;
Rachel: x2 = 7(t – 4), y2 = 4
x
100
y
5
4
3
2
1
x
100
y
5
4
3
2
1
D) Bus: x1 = 3t2, y1 = 2;
Rachel: x2 = 7
5(t – 4), y2 = 4
x
100
y
5
4
3
2
1
x
100
y
5
4
3
2
1
Page 100
6) Car A (travelling north at 60 mph) and car B (traveling west at 30 mph) are heading toward the same
intersection. Car A is 4 miles from the intersection when car B is 6 miles from the intersection. Find
parametric equations that describe the motion of cars A and B.
6 mi car B 30 mph
4 mi
car A 60 mph
A) Car A: x = 0, y = 60t – 4; Car B: x =6–30t, y =0
B) Car A: x = –60t + 4
,
y = 0; Car B: x =6–30t, y =0
C) Car A: x = 0, y = 30t – 6; Car B: x =60t –4
,
y =0
D) Car A: x = 30t – 6
,
y = 0; Car B: x =0, y =4–60t
7) Car A (travelling north at 70 mph) and car B (traveling west at 50 mph) are heading toward the same
intersection. Car A is 3 miles from the intersection when car B is 4 miles from the intersection. . Find a
formula for the distance between the cars as a function of time, using the parametric equations that
describe the motion of cars A and B. Using a graphing utility, find the minimum distance between the
cars. When are the cars closest?
4 mi car B 50 mph
3 mi
car A 70 mph
A) d = (70t
– 3)2 + (4 – 50t)2; 1.51 mi; 3.32 min
B) d = (70t
– 3)2 + (4 – 50t)2; 1.51 mi; 0.06 min
C) d = (70t
– 3)2 + (4 – 50t)2; 281.21 mi; 3.32 min
D) d = (70t
+ 3)2 + (4 – 50t)2; 1.51 mi; 3.32 h
5 Find Parametric Equations for Curves Defined by Rectangular Equations
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find two sets of parametric equations for the given rectangular equation.
1) y = 4x + 7
A) x = t, y = 4t + 7; x = t
4, y = t + 7B)x =t, y =4t +7; x = 4t, y = t + 7
C) x = t, y = 4t + 7; x = t, y = t
4 + 7D)x = 4t, y = t + 7; x = t
4, y = t + 7
Page 101
Find parametric equations for the rectangular equation.
2) y = 2x + 2
A) x = t, y = 2t + 2; 0 ≤ t
<
∞B) y =2t, 2x =t –2; 0 ≤ t
<
∞
C) x = t
2, y = t + 1; 0 ≤ t < ∞D) x = t, y = 2t2 + 2; 0 ≤ t < ∞
3) y = x4 – 2
A) x = t, y = t4 – 2; 0 ≤ t < ∞B) x = t2, y = t2 – 2; 0 ≤ t < ∞
C) x = t2, y = t4 – 2; 0 ≤ t < ∞D) x = t, y = t2 – 2; 0 ≤ t < ∞
4) y = 3x2 + 6
A) x = t; y = 3t2 + 6; 0 ≤ t < ∞B) y = t; x = 3t2 + 6; 0 ≤ t < ∞
C) x = t2; y = 3t + 6; 0 ≤ t < ∞D) x = t; y = 3t + 6; t ≥ 0
Find the parametric equations that define the curve shown.
5)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(3, –1)
(6, –4)
x
–10–8-6-4-2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
(3, –1)
(6, –4)
A) x = t + 3
,
y = –t – 1; 0 ≤ t ≤ 3B)x = –t+3
,
y = –t – 1; 0 ≤ t ≤ 2
C) x = –t + 3
,
y = t – 1; 0 ≤ t ≤ 3D)x =t+3
,
y = –t –1; 0 ≤ t ≤ 2
Page 102
6)
x
-8 -6 -4 -2 2 4 6 8
y
8
6
4
2
-2
-4
-6
-8
(-3, 0) (3, 0)
x
-8 -6 -4 -2 2 4 6 8
y
8
6
4
2
-2
-4
-6
-8
(-3, 0) (3, 0)
A) x = 3sin π
2 (t – 1) , y = –5 cos π
2 (t – 1) ; 0 ≤ t ≤ 4
B) x = 3sin π
2 (t – 1) , y = –5 cos π
2 (t – 1) ; 0 ≤ t ≤ 3
C) x = 5sin π
2 (t – 1) , y = –3 cos π
2 (t – 1) ; 0 ≤ t ≤ 4
D) x = –5sin π
2 (t – 1) , y = 3 cos π
2 (t – 1) ; 0 ≤ t ≤ 3
Solve the problem.
7) Find parametric equations for an object that moves along the ellipse x2
9 + y2
16 = 1 with the motion
described.
The motion begins at (0, 4), is clockwise, and requires 3 seconds for a complete revolution.
A) x = 3 sin (2
3
πt), y = 4 cos (2
3
πt), 0 ≤ t ≤ 3B)x = 3 cos (2
3
πt), y = 4 sin (2
3
πt), 0 ≤ t ≤ 3
C) x = 3 sin (2
3
πt), y = –4 cos (2
3
πt), 0 ≤ t ≤ 3D)x =3 sin 3πt
,
y = 4 cos 3πt
,
0 ≤t ≤3
Page 103
Ch. 11 Analytic Geometry
Answer Key
11.1 Conics
11.2 The Parabola
Page 104
11.3 The Ellipse
Page 105
11.4 The Hyperbola
1 Analyze Hyperbolas with Center at the Origin
Page 106
11.5 Rotation of Axes; General Form of a Conic
11.6 Polar Equations of Conics
Page 107
11.7 Plane Curves and Parametric Equations
Page 108
Page 109