Ch. 7 Analytic Geometry
7.1 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
7.2 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 = 9x
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 = –5x
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 = 5y
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 = –5y
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 (–20
,
0); directrix the line x =20
A) y2 = –80x B) y2 = 80x C) y2 = –20x D) x2 = –80y
6) Focus at (0, –5); directrix the line y =5
A) x2 = –20y B) x2 = 20y C) y2 = –5x D) y2 = –20x
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 (8
,
3)
A) y2 = 9
8xB)y
2 = 9
32xC)x
2 = 9
8yD)x
2 = 9
32y
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
-10 -5 5 10
y
10
5
-5
-10
-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 = 16x
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) 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
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: (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
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 7
Graph the equation.
14) y2 = 6x
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Page 8
15) y2 = –9x
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Page 9
16) x2 = 16y
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-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)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-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)
(2, 4)
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, 4)
A) y2 = 8x B) x2 = –8y C) x2 = 8y D) y2 = –8x
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 = 7(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 + 2)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 – 2)2 = –5(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 (2
,
5); focus at (4
,
5)
A) (y – 5)2 = 8(x – 2) B) (y – 5)2 = –8(x – 2)
C) (x – 5)2 = –4(y – 5) D) (x – 5)2 = 4(y – 5)
6) Vertex at (1
,
4); focus at (1
,
3)
A) (x – 1)2 = –4(y – 4) B) (x – 1)2 = 4(y – 4)
C) (y – 4)2 = –8(x – 1) D) (y – 4)2 = 8(x – 1)
7) Vertex at (8
,
–2); focus at (8
,
–3)
A) (x – 8)2 = –4(y + 2) B) (x – 8)2 = 4(y + 2)
C) (y – 2)2 = –20(x + 8) D) (y – 2)2 = 20(x + 8)
8) Vertex at (5
,
–6); focus at (1
,
–6)
A) (y + 6)2 = –16(x – 5) B) (y + 6)2 = 16(x – 5)
C) (x + 5)2 = 20(y – 6) D) (x + 5)2 = –20(y – 6)
9) Focus at (–2
,
–4); directrix the line y =0
A) (x + 2)2 = –8(y + 2) B) (x – 2)2 = –8(y + 2)
C) (x + 2)2 = –8(y – 2) D) (x – 2)2 = –8(y – 2)
Page 15
Find the vertex, focus, and directrix of the parabola with the given equation.
10) (y + 1)2 = 16(x + 4)
A) vertex: (–4
,
–1)
focus: (0, –1)
directrix: x = –8
B) vertex: (4
,
1)
focus: (8, 1)
directrix: x = 0
C) vertex: (–1
,
–4)
focus: (3, –4)
directrix: x = –5
D) vertex: (–4
,
–1)
focus: (–8, –1)
directrix: x = 0
11) (y – 3)2 = –4(x + 2)
A) vertex: (–2
,
3)
focus: (–3, 3)
directrix: x = –1
B) vertex: (2
,
–3)
focus: (1, –3)
directrix: x = 3
C) vertex: (3
,
–2)
focus: (2, –2)
directrix: x = 4
D) vertex: (–2
,
3)
focus: (–1, 3)
directrix: x = –3
12) (x + 3)2 = 12(y + 1)
A) vertex: (–3
,
–1)
focus: (–3, 2)
directrix: y = –4
B) vertex: (3
,
1)
focus: (3, 4)
directrix: y = –2
C) vertex: (–1
,
–3)
focus: (–1, 0)
directrix: y = –6
D) vertex: (–3
,
–1)
focus: (–3, –4)
directrix: x = 2
13) (x + 4)2 = –4(y – 3)
A) vertex: (–4
,
3)
focus: (–4, 2)
directrix: y = 4
B) vertex: (4
,
–3)
focus: (4, –4)
directrix: y = –2
C) vertex: (3
,
–4)
focus: (3, –5)
directrix: y = –3
D) vertex: (–4
,
3)
focus: (–4, 4)
directrix: x = 2
Find the vertex, focus, and directrix of the parabola. Graph the equation.
14) (y – 2)2 = –4(x – 3)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (3
,
2)
focus: (2, 2)
directrix: x = 4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) vertex: (–2
,
–3)
focus: (–3, –3)
directrix: x = –1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 16
C) vertex: (3
,
2)
focus: (3, 1)
directrix: y = 3
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (–3
,
–2)
focus: (–3, –3)
directrix: y = –1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
15) (x + 1)2 = –8(y + 3)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) 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
B) 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
Page 17
C) vertex: (–1
,
–3)
focus: (–3, –3)
directrix: x = 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, 3)
directrix: x = 3
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
16) x2 – 6x = 8y – 49
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (3
,
5)
focus: (3, 7)
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: (3
,
5)
focus: (3, 3)
directrix: y = 7
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 18
C) vertex: (3
,
5)
focus: (5, 5)
directrix: x = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (3
,
5)
focus: (1, 5)
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 + 12y = 4x – 16
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) vertex: (–5
,
–6)
focus: (–4, –6)
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: (–5
,
–6)
focus: (–6, –6)
directrix: x = –4
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
Page 19
C) vertex: (–5
,
–6)
focus: (–5, –5)
directrix: y = –7
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) vertex: (–5
,
–6)
focus: (–5, –7)
directrix: y = –5
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 = 7(x + 2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Page 20
19) (y – 1)2 = –6(x + 2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Page 21
20) (x + 2)2 = 6(y – 2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Page 22
21) (x + 1)2 = –6(y + 2)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
B)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
C)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
D)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
Write an equation for the parabola.
22)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
points: (3, –2), (5, 2)
A) (y + 2)2 = 8(x – 3) B) (y + 3)2 = 8(x – 2) C) (x + 2)2 = 8(y – 3) D) (x – 2)2 = 8(y + 3)
Page 23
23)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
points: (4, –2), (8, 0)
A) (x – 4)2 = 8(y + 2) B) (x – 2)2 = 8(y + 4) C) (y – 4)2 = 8(x + 2) D) (x + 2)2 = 8(y – 4)
24)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
points: (–1, 2), (1, 1)
A) (x + 1)2 = –4(y – 2) B) (x + 2)2 = –4(y – 1)
C) (y + 1)2 = –4(x – 2) D) (x – 2)2 = 4(y + 1)
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 20 inches
across at its opening and is 5 feet deep, where will the light be concentrated?
A) 0.4 in. from the vertex B) 5 in. from the vertex
C) 1.3 in. from the vertex D) 0.6 in. from the vertex
2) A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base
along the axis of symmetry and the opening is 10 feet across, how deep should the searchlight be?
A) 3.1 ft B) 0.2 ft C) 12.5 ft D) 6.3 ft
3) A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 188 feet and a maximum
height of 30 feet. Find the height of the arch at 25 feet from its center.
A) 27.9 ft B) 0.5 ft C) 8.5 ft D) 65.3 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 35 feet and the distance across the top of the mirror is 78 inches, how deep is the mirror in the
center?
A) 507
560 in. B) 1521
140 in. C) 169
2240 in. D) 1225
156 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 9 inches tall and stand 60 inches apart. Find the vertical distance
from the roadway to the cable at a point on the road 9 inches from the lowest point of the cable.
A) 0.81 in. B) 3.24 in. C) 1.01 in. D) 0.61 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 6.25 inches tall and stand 50 inches apart. At some point along the
road from the lowest point of the cable, the cable is 0.56 inches above the roadway. Find the distance
between that point and the base of the nearest tower.
A) 17.5 in. B) 7.3 in. C) 17.7 in. D) 7.7 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 50 inches apart. At a point between the towers and 17.5 inches along
the road from the base of one tower, the cable is 0.56 inches above the roadway. Find the height of the
towers.
A) 6.25 in. B) 6.75 in. C) 5.75 in. D) 8.25 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
7.3 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)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) x2
4 + y2
9 = 1B)
y2
9 – x2
4 = 1C)
x2
4 – y2
9 = 1D)
x2
9 + y2
4 = 1
2)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) x2
49 + y2
9 = 1B)
y2
9 – x2
49 = 1C)
– y2
9 + x2
49 = 1D)
x2
9 + y2
49 = 1
Find the foci and vertices of the ellipse.
3) x2
81 + y2
16 = 1
A) foci at (– 65, 0) and ( 65, 0)
vertices at (–9, 0), (9, 0)
B) foci at (0, – 65) and (0, 65)
vertices at (0, –9), (0, 9)
C) foci at (–9
,
0) and (9
,
0)
vertices at (–81, 0), (81, 0)
D) foci at (0, –4) and (0, 4)
vertices at (0, –16), (0, 16)
Page 26
4) x2
9 + y2
64 = 1
A) foci at (0, – 55) and (0, 55)
vertices at (0, –8), (0, 8)
B) foci at (– 55, 0) and ( 55, 0)
vertices at (–8, 0), (8, 0)
C) foci at (0, –8) and (0, 8)
vertices at (0, –64), (0, 64)
D) foci at (0, 8) and (3
,
0)
vertices at (0, 64), (9, 0)
5) 25x2 + 81y2 = 2025
A) foci at (– 214
, 0) and (2 14, 0)
vertices at (–9, 0), (9, 0)
B) foci at (0, – 214) and (0, 2 14)
vertices at (0, –9), (0, 9)
C) foci at (–9
,
0) and (9
,
0)
vertices at (–81, 0), (81, 0)
D) foci at (0, –5) and (0, 5)
vertices at (0, –25), (0, 25)
6) 49x2 + 16y2 = 784
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) and (16, 0)
Find an equation for the ellipse.
7) Center at (0, 0); focus at (4
,
0); vertex at (5
,
0)
A) x2
25 + y2
9 = 1B)
x2
9 + y2
25 = 1C)
x2
16 + y2
9 = 1D)
x2
16 + y2
25 = 1
8) Center at (0, 0); focus at (–2
,
0); vertex at (6
,
0)
A) x2
36 + y2
32 = 1B)
x2
32 + y2
36 = 1C)
x2
4 + y2
32 = 1D)
x2
4 + y2
36 = 1
9) Center at (0, 0); focus at (–7
,
0); vertex at (8
,
0)
A) x2
64 + y2
15 = 1B)
x2
15 + y2
64 = 1C)
x2
49 + y2
15 = 1D)
x2
49 + y2
64 = 1
10) Vertices at (0, ±8); c = 6
A) x2
28 + y2
64 = 1B)
x2
64 + y2
28 = 1C)
x2
36 + y2
28 = 1D)
x2
36 + y2
64 = 1
11) Center at (0, 0); focus at (0, –2); vertex at (0, 7)
A) x2
45 + y2
49 = 1B)
x2
49 + y2
45 = 1C)
x2
4 + y2
45 = 1D)
x2
4 + y2
49 = 1
12) Foci at (0, ±3); a = 6
A) x2
27 + y2
36 = 1B)
x2
36 + y2
27 = 1C)
x2
9 + y2
27 = 1D)
x2
9 + y2
36 = 1
13) Focus at (–6
,
0); vertices at (±8
,
0)
A) x2
64 + y2
28 = 1B)
x2
28 + y2
64 = 1C)
x2
36 + y2
28 = 1D)
x2
36 + y2
64 = 1
Page 27
14) Focus at (0, –2); vertices at (0, ±8)
A) x2
60 + y2
64 = 1B)
x2
64 + y2
60 = 1C)
x2
4 + y2
60 = 1D)
x2
4 + y2
64 = 1
15) Foci at (±3
,
0); x–intercepts are ±5
A) x2
25 + y2
16 = 1B) x2
16 + y2
25 = 1C)
x2
9 + y2
16 = 1D)
x2
9 + y2
25 = 1
16) Foci at (0, ±4); y–intercepts are ±6
A) x2
20 + y2
36 = 1B) x2
36 + y2
20 = 1C)
x2
16 + y2
20 = 1D)
x2
16 + y2
36 = 1
17) Center (0, 0); major axis horizontal with length 14; length of minor axis is 8
A) x2
49 + y2
16 = 1B)
x2
16 + y2
49 = 1C)
x2
14 + y2
16 = 1D)
x2
196 + y2
64 = 1
18) Center (0, 0); major axis vertical with length 14; length of minor axis is 8
A) x2
16 + y2
49 = 1B)
x2
49 + y2
16 = 1C)
x2
8 + y2
49 = 1D)
x2
64 + y2
196 = 1
19) Foci at (0, ±3); length of the major axis is 10
A) x2
16 + y2
25 = 1B)
x2
25 + y2
16 = 1C)
x2
16 + y2
5 = 1D)
x2
25 + y2
5 = 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 + 9y2 = 36
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
A) 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
B) foci at (0, 5) and (0, –5)
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) foci at ( 13, 0) and (–13, 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 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
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
16 + (y + 2)2
9 = 1B)
(x + 2)2
16 + (y + 1)2
9 = 1
C) (x – 1)2
16 + (y – 2)2
9 = 1D)
(x + 1)2
9 + (y + 2)2
16 = 1
Find the center, foci, and vertices of the ellipse.
2) (x + 2)2
16 + (y – 3)2
9 = 1
A) center at (–2
,
3)
foci at (–2 + 7
, 3), (–2 – 7, 3)
vertices at (–6, 3), (2, 3)
B) center at (3
,
–2)
foci at (3 + 7
, –2), (3 – 7, –2)
vertices at (–6, 3), (2, 3)
C) center at (–2
,
3)
foci at (– 7
, 3), ( 7, 3)
vertices at (4, 3), (–4, 3)
D) center at (–2
,
3)
foci at (–2 + 7
, –2), (–2 – 7, –2)
vertices at (4, 3), (–4, 3)
3) 36(x – 3)2 + 9(y + 2)2 = 324
A) center at (3
,
–2)
foci at (3, –2 – 33
), (3, –2 + 33)
vertices at (3, 4), (3, –8)
B) center at (–2
,
3)
foci at (–2, 3 – 33
), (–2, 3 + 33)
vertices at (–2, 4), (–2, –8)
C) center at (–3
,
–2)
foci at (–3, –2 – 33
), (–3, –2 + 33)
vertices at (–3, 4), (–3, –8)
D) center at (4
,
–2)
foci at (4, –2 – 33
), (4, –2 + 33)
vertices at (4, 4), (4, –8)
Page 33
4) 2x2 + 5y2 – 24x + 70y + 307 = 0
A) (x – 6)2
5 + (y + 7)2
2 = 1
center: (6, –7); foci: (7.7, –7), (4.3, –7); vertices: (8.2, –7), (3.8, –7)
B) (x – 6)2
2 + (y + 7)2
5 = 1
center: (6, –7); foci: (7.7, –7), (4.3, –7); vertices: (8.2, –7), (3.8, –7)
C) (x – 6)2
5 + (y + 7)2
2 = 1
center: (–6, 7); foci: (–4.3, 7), (–7.7, 7); vertices: (–8.2, 7), (–3.8, 7)
D) (x – 6)2
2 + (y + 7)2
5 = 1
center: (–6, 7); foci: (–4.3, 7), (–7.7, 7); vertices: (–8.2, 7), (–3.8, 7)
5) 9x2 + y2 – 36x + 27 = 0
A) (x – 2)2 + y2
9 = 1
center: (2, 0); foci: (2, 2 2), (2, –22); vertices:(2, 3), (2, –3)
B) x2
9 + (y – 2)2 = 1
center: (2, 0); foci: (2, 2 2), (2, –22); vertices:(2, 3), (2, –3)
C) (x – 3)2 + y2
4 = 1
center: (3, 0); foci: (3, 3), (3, –3); vertices:(3, 2), (3, –2)
D) x2
4 + (y – 3)2 = 1
center: (3, 0); foci: (3, 3), (3, –22); vertices:(3, 2),
(3, –2)
Page 34
Graph the equation.
6) (x – 1)2
9 + (y + 1)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 + 1)2
4 + (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
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 + 16(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
Page 37
9) 16(x – 2)2 + 9(y – 1)2 = 144
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 (3
,
5); focus at (10
,
5); vertex at (12
,
5)
A) (x – 3)2
81 + (y – 5)2
32 = 1B)
(x + 3)2
81 + (y + 5)2
32 = 1
C) (x – 3)2
144 + (y + 5)2
11 = 2D)
(x + 3)2
49 – (y – 5)2
25 = 1
11) Vertices at (0
,
6) and (10
,
6); focus at (8
,
6)
A) (x – 5)2
25 + (y – 6)2
16 = 1B)
(x – 6)2
16 + (y – 5)2
15 = 1
C) (x + 5)2
9 + (y + 6)2
16 = 1D)
(x – 5)2
49 – (y + 6)2
19 = 1
Page 38
Find an equation for the ellipse described. Graph the equation.
12) Foci at (2
,
5) and (2
,
–1); length of major axis is 10
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) (y – 2)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 – 2)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 – 2)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 – 2)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 (–6
,
–3) and (0
,
–3); length of major axis is 10
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) (x + 3)2
25 + (y + 3)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 – 4)2
25 + (x + 3)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 + 4)2
25 + (x – 3)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 – 4)2
25 + (y – 3)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
-10 -5 5 10
y
10
5
-5
-10
-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 (–2
,
3) and (–8
,
3); vertex at (–9
,
3)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) (x + 5)2
16 + (y – 3)2
7 = 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
7 + (y – 3)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 – 3)2
16 + (y + 5)2
7 = 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
7 + (y + 5)2
16 = 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 (–3
,
4); focus at (–6
,
4); contains the point (–7
,
4)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) (x + 3)2
16 + (y – 4)2
7 = 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
7 + (y – 4)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 + 4)2
16 + (y – 3)2
7 = 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
7 + (y – 3)2
16 = 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 = – 9
– 4x2
y
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 120 feet. The height of the arch 29 feet
from the center is to be 9 feet. Find the height of the arch at its center.
A) 10.28 ft B) 9.27 ft C) 29.33 ft D) 18.62 ft
3) An arch for a bridge over a highway is in the form of a semiellipse. The top of the arch is 30 feet above
ground (the major axis). What should the span of the bridge be (the length of its minor axis) if the height
29 feet from the center is to be 10 feet above ground?
A) 61.52 ft B) 30.76 ft C) 174 ft D) 78.11 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 540.9 million miles and a mean
distance of 543 million miles, write an equation for the orbit of the planet around the sun.
A) x2
5432 + y2
542.9962 = 1B)
x2
543.0042 + y2
5432 = 1
C) x2
5432 + y2
2.12 = 1D)
x2
5432 + y2
540.92 = 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
7.4 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
16 – 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
9 – 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
4x
A) y2
36 – x2
64 = 1B)
y2
64 – x2
36 = 1C)
y2
36 – x2
16 = 1D)
y2
16 – x2
9 = 1
4) Vertices at (±2
,
0); foci at (±8
,
0)
A) x2
4 – y2
60 = 1B)
x2
60 – y2
4 = 1C)
x2
4 – y2
64 = 1D)
x2
64 – y2
4 = 1
Page 47
Find an equation for the hyperbola described. Graph the equation.
5) Center at (0, 0); focus at (2 17, 0); vertex at (2, 0)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) x2
4 – y2
64 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
B) x2
64 – y2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
C) y2
64 – x2
4 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) y2
4 – x2
64 = 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, 4); focus at (0, 65)
-10 -5 5 10
y
10
5
-5
-10
-10 -5 5 10
y
10
5
-5
-10
A) y2
16 – 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
16 = 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
16 = 1
x
-10 -5 5 10
y
10
5
-5
-10
x
-10 -5 5 10
y
10
5
-5
-10
D) x2
16 – 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
25 – y2
100 = 1
A) center at (0, 0)
transverse axis is x–axis
vertices at (–5, 0) and (5, 0)
foci at (– 55
, 0) and (5 5, 0)
asymptotes of y = – 2 and y = 2
B) center at (0, 0)
transverse axis is x–axis
vertices at (–10, 0) and (10, 0)
foci at (– 55, 0) and (5 5, 0)
asymptotes of y = – 2 and y = 2
C) center at (0, 0)
transverse axis is y–axis
vertices at (0, –5) and (0, 5)
foci at (– 55
, 0) and (5 5, 0)
asymptotes of y = – 2 and y = 2
D) center at (0, 0)
transverse axis is x–axis
vertices at (–5, 0) and (5, 0)
foci at (–10, 0) and (10, 0)
asymptotes of y = – 2 and y = 2
8) 25y2 – 100x2 = 2500
A) center at (0, 0)
transverse axis is y–axis
vertices at (0, –10) and (0, 10)
foci at (0, – 55
) and (0, 5 5)
asymptotes of y = – 2 and y = 2
B) center at (0, 0)
transverse axis is x–axis
vertices: (–5, 0), (5, 0)
foci: (– 55, 0) , (5 5, 0)
asymptotes of y = – 2 and y = 2
C) center at (0, 0)
transverse axis is y–axis
vertices: (0, –10), (0, 10)
foci: (– 55
, 0) , (5 5, 0)
asymptotes of y = – 2 and y = 2
D) center at (0, 0)
transverse axis is x–axis
vertices: (–10, 0), (10, 0)
foci: (–5, 0), (5, 0)
asymptotes of y = – 2 and y = 2
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
16 – y2
25 = 1B)
y2
16 – x2
25 = 1C)
x2
25 – y2
16 = 1D)
y2
25 – x2
16 = 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
25 = 1B)
x2
9 – y2
25 = 1C)
x2
25 – y2
9 = 1D)
y2
25 – 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
4 – x2
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 53
13) 25x2 – 9y2 = 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 54
14) 25y2 – 4x2 = 100
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) 25x2 = 9y2 + 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 56
16) 9y2 = 4x2 + 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
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
4 – y2
9 = 1
A) y = 3
2x and y = – 3
2xB)y
= 2
3x and y = – 2
3x
C) y = 9
4x and y = – 9
4xD)y
= 4
9x and y = – 4
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 – 1)2
16 – (y + 2)2
9 = 1
A) y + 2 = 3
4(x – 1) and y + 2 = – 3
4(x – 1) B) y + 2 = 4
3(x – 1) and y + 2 = – 4
3(x – 1)
C) y = 3
4(x – 1) and y = – 3
4(x – 1) D) y – 1 = 3
4(x + 2) and y – 1 = – 3
4(x + 2)
4) x2 – y2 – 4x + 8y – 21 = 0
A) y – 4 = (x – 2) and y – 4 = – (x – 2) B) y – 4 = 1
3(x – 2) and y – 4 = – 1
3(x – 2)
C) y + 2 = (x + 4) and y + 2 = – (x + 4) D) y –2=(x –4) and y – 2 = – (x –4)
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 (8
,
7); focus at (2
,
7); vertex at (7
,
7)
A) (x – 8)2 – (y – 7)2
35 = 1B)
(x – 8)2
35 – (y – 7)2 = 1
C) (x – 7)2 – (y – 8)2
35 = 1D)
(x – 7)2
35 – (y – 8)2 = 1
3) Vertices at (0, ±4); asymptotes at y = ± 2
3x
A) y2
16 – x2
36 = 1B)
y2
36 – x2
16 = 1C) y2
16 – x2
9 = 1D)
y2
9 – x2
4 = 1
4) Vertices at (±6
,
0); foci at (±10
,
0)
A) x2
36 – y2
64 = 1B)
x2
64 – y2
36 = 1C)
x2
36 – y2
100 = 1D)
x2
100 – y2
36 = 1
Page 58
Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
5) (x – 3)2
9 – (y + 3)2
25 = 1
A) center at (3
,
–3)
transverse axis is parallel to x–axis
vertices at (0, –3) and (6, –3)
foci at (3 – 34, –3) and (3 + 34, –3)
asymptotes of y + 3 = – 5
3(x – 3) and y + 3 = 5
3(x – 3)
B) center at (–3
,
3)
transverse axis is parallel to x–axis
vertices at (–6, 3) and (0, 3)
foci at (–3 – 34, 3) and (–3 + 34, 3)
asymptotes of y – 3 = – 5
3(x + 3) and y – 3 = 5
3(x + 3)
C) center at (3
,
–3)
transverse axis is parallel to y–axis
vertices at (3, –6) and (3, 0)
foci at (3, –3 – 34) and (3, –3 + 34)
asymptotes of y – 3 = – 3
5(x + 3) and y – 3 = 3
5(x + 3)
D) center at (3
,
–3)
transverse axis is parallel to x–axis
vertices at (–2, –3) and (8, –3)
foci at (3 – 34, –3) and (3 + 34, –3)
asymptotes of y + 3 = – 3
5(x – 3) and y + 3 = 3
5(x – 3)
Page 59
6) (x – 2)2 – 16(y + 2)2 = 16
A) center at (2
,
–2)
transverse axis is parallel to x–axis
vertices at (–2, –2) and (6, –2)
foci at (2 – 17, –2) and (2 + 17, –2)
asymptotes of y + 2 = – 1
4(x – 2) and y + 2 = 1
4(x – 2)
B) center at (–2
,
2)
transverse axis is parallel to x–axis
vertices at (–6, 2) and (2, 2)
foci at (–2 – 17, 2) and (–2 + 17, 2)
asymptotes of y – 2 = – 1
4(x + 2) and y – 2 = 1
4(x + 2)
C) center at (2
,
–2)
transverse axis is parallel to y–axis
vertices at (2, –6) and (2, 2),
foci at (2, –2 – 17) and (2, –2 + 17),
asymptotes of y – 2 = – 4(x + 2) and y – 2 = 4(x + 2)
D) center at (2
,
–2)
transverse axis is parallel to x–axis
vertices at (1, –2) and (3, –2)
foci at (2 – 17, –2) and (2 + 17, –2)
asymptotes of y + 2 = – 4(x – 2) and y + 2 = 4(x – 2)
7) x2 – 16y2 + 6x – 32y – 23 = 0
A) center at (–3
,
–1)
transverse axis is parallel to x–axis
vertices at (–7, –1) and (1, –1)
foci at (–3 – 17, –1) and (–3 + 17, –1)
asymptotes of y + 1 = – 1
4(x + 3) and y + 1 = 1
4(x + 3)
B) center at (–1
,
–3)
transverse axis is parallel to x–axis
vertices at (–5, –3) and (3, –3)
foci at (–1 – 17, –3) and (–1 + 17, –3)
asymptotes of y + 3 = – 1
4(x + 1) and y + 3 = 1
4(x + 1)
C) center at (–3
,
–1)
transverse axis is parallel to y–axis
vertices at (–3, –5) and (–3, 3)
foci at (–3, –1 – 17) and (–3, –1 + 17)
asymptotes of y – 1 = – 4(x – 3) and y – 1 = 4(x – 3)
D) center at (–3
,
–1)
transverse axis is parallel to x–axis
vertices at (–4, –1) and (–2, –1)
foci at (–3 – 17, –1) and (–3 + 17, –1)
asymptotes of y + 1 = – 4(x + 3) and y + 1 = 4(x + 3)
Page 60
Graph the hyperbola.
8) (x + 2)2
9 – (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
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
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 62
10) (x – 2)2 – 9(y – 1)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 63
11) (y – 3)2 – 9(x + 1)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 = –6 + x2
-10 -5 5 10
y
10
5
-5
-10
-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, 300 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) 681.82 ft B) 3947.57 ft C) 1367.48 ft D) 1401.74 ft
Page 65
3) The roof of a building is in the shape of the hyperbola y2 – x2 = 32, where x and y are in meters. Refer to
the figure and determine the height h of the outside walls.
a = b = 4 m
A) 6.9 m B) 48 m C) –16 m D) 28 m
4) The roof of a building is in the shape of the hyperbola y2 – x2 = 37, 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) 15.9 m B) 63 m C) 7.95 m D) 11.7 m
5) A comet follows the hyperbolic path described by x2
8 – 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) 2.2 million B) 5 million C) 2.8 million D) 25 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
3x 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
3x
A) y2
16 – x2
9
4
= 1B)
x2
16 – y2
16
3
2 = 1C)
y2
9
4
– x2
16 = 1D)
x2
16
3
2 – y2
16 = 1
Page 67
Ch. 7 Analytic Geometry
Answer Key
7.1 Conics
1 Know the Names of the Conics
7.2 The Parabola
1 Analyze Parabolas with Vertex at the Origin
2 Analyze Parabolas with Vertex at (h, k)
Page 68
3 Solve Applied Problems Involving Parabolas
7.3 The Ellipse
1 Analyze Ellipses with Center at the Origin
2 Analyze Ellipses with Center at (h, k)
Page 69
3 Solve Applied Problems Involving Ellipses
7.4 The Hyperbola
1 Analyze Hyperbolas with Center at the Origin
2 Find the Asymptotes of a Hyperbola
3 Analyze Hyperbolas with Center at (h, k)
4 Solve Applied Problems Involving Hyperbolas
Page 70
Page 71