Question 1

Find an equation for the ellipse described.
-Focus at (-2, 0); vertices at (-8, 0)and (8, 0)

Accepted Answer

A)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 60 } = 1$ 
B)  $\frac { x ^ { 2 } } { 60 } + \frac { y ^ { 2 } } { 64 } = 1$ 
C)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 64 } = 1$ 
D)  $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 60 } = 1$ 
A)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 60 } = 1$ 
B)  $\frac { x ^ { 2 } } { 60 } + \frac { y ^ { 2 } } { 64 } = 1$ 
C)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 64 } = 1$ 
D)  $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 60 } = 1$

Question 2

Find an equation for the ellipse described.
-Foci at (-3, 4)and (-3, -2); length of major axis is 10

Accepted Answer

A)  $\frac { ( y - 1 ) ^ { 2 } } { 25 } + \frac { ( x - 3 ) ^ { 2 } } { 16 } = 1$ 
B)  $\frac { ( x - 1 ) ^ { 2 } } { 25 } + \frac { ( y - 3 ) ^ { 2 } } { 16 } = 1$ 
C)  $\frac { ( y - 1 ) ^ { 2 } } { 25 } + \frac { ( x + 3 ) ^ { 2 } } { 16 } = 1$ 
D)  $\frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y - 3 ) ^ { 2 } } { 25 } = 1$ 
A)  $\frac { ( y - 1 ) ^ { 2 } } { 25 } + \frac { ( x - 3 ) ^ { 2 } } { 16 } = 1$ 
B)  $\frac { ( x - 1 ) ^ { 2 } } { 25 } + \frac { ( y - 3 ) ^ { 2 } } { 16 } = 1$ 
C)  $\frac { ( y - 1 ) ^ { 2 } } { 25 } + \frac { ( x + 3 ) ^ { 2 } } { 16 } = 1$ 
D)  $\frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y - 3 ) ^ { 2 } } { 25 } = 1$

Question 3

Graph the ellipse and locate the foci.
-$$9 x ^ { 2 } + 4 y ^ { 2 } = 36$$

Accepted Answer

A)  foci at $( 0 , \sqrt { 5 } )$ and $( 0 , - \sqrt { 5 } )$
  
B)  foci at $( 2 \sqrt { 3 } , 0 )$ and $( - 2 \sqrt { 3 } , 0 )$
  
C)  foci at $( \sqrt { 5 } , 0 )$ and $( - \sqrt { 5 } , 0 )$
  
D)  foci at $( \sqrt { 13 } , 0 )$ and $( - \sqrt { 13 } , 0 )$
  
A)  foci at $( 0 , \sqrt { 5 } )$ and $( 0 , - \sqrt { 5 } )$
  
B)  foci at $( 2 \sqrt { 3 } , 0 )$ and $( - 2 \sqrt { 3 } , 0 )$
  
C)  foci at $( \sqrt { 5 } , 0 )$ and $( - \sqrt { 5 } , 0 )$
  
D)  foci at $( \sqrt { 13 } , 0 )$ and $( - \sqrt { 13 } , 0 )$

Question 4

Graph the ellipse and locate the foci.
-$$\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$$

Accepted Answer

A)  foci at $( 0,2 \sqrt { 3 } )$ and $( 0 , - 2 \sqrt { 3 } )$
  
B)  foci at $( \sqrt { 21 } , 0 )$ and $( - \sqrt { 21 } , 0 )$
  
C)  foci at $( 2 \sqrt { 3 } , 0 )$ and $( - 2 \sqrt { 3 } , 0 )$
  
D)  foci at $( 2 \sqrt { 5 } , 0 )$ and $( - 2 \sqrt { 5 } , 0 )$
  
A)  foci at $( 0,2 \sqrt { 3 } )$ and $( 0 , - 2 \sqrt { 3 } )$
  
B)  foci at $( \sqrt { 21 } , 0 )$ and $( - \sqrt { 21 } , 0 )$
  
C)  foci at $( 2 \sqrt { 3 } , 0 )$ and $( - 2 \sqrt { 3 } , 0 )$
  
D)  foci at $( 2 \sqrt { 5 } , 0 )$ and $( - 2 \sqrt { 5 } , 0 )$

Question 5

Graph the equation.
-$$( x + 1 ) ^ { 2 } = - 8 ( y + 2 )$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 6

Find an equation for the ellipse described.
-Center (0, 0); major axis horizontal with length 8; length of minor axis is 4

Accepted Answer

A)  $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$ 
C)  $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$ 
D)  $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$ 
A)  $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$ 
C)  $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$ 
D)  $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$

Question 7

Find an equation for the ellipse described.
-Vertices at (-6, 2)and (14, 2); focus at (12, 2)

Accepted Answer

A)  $\frac { ( x - 2 ) ^ { 2 } } { 81 } + \frac { ( y - 4 ) ^ { 2 } } { 35 } = 1$ 
B)  $\frac { ( x - 4 ) ^ { 2 } } { 144 } - \frac { ( y + 2 ) ^ { 2 } } { 44 } = 1$ 
C)  $\frac { ( x - 4 ) ^ { 2 } } { 100 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$ 
D)  $\frac { ( x + 4 ) ^ { 2 } } { 64 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$ 
A)  $\frac { ( x - 2 ) ^ { 2 } } { 81 } + \frac { ( y - 4 ) ^ { 2 } } { 35 } = 1$ 
B)  $\frac { ( x - 4 ) ^ { 2 } } { 144 } - \frac { ( y + 2 ) ^ { 2 } } { 44 } = 1$ 
C)  $\frac { ( x - 4 ) ^ { 2 } } { 100 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$ 
D)  $\frac { ( x + 4 ) ^ { 2 } } { 64 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$

Question 8

Find the equation in standard form of the parabola described.
-The vertex has coordinates (0, 0), and the focus has coordinates (6, 0).

Accepted Answer

A)  $y ^ { 2 } = 24 x$ 
B)  $x ^ { 2 } = 6 y$ 
C)  $x ^ { 2 } = 24 y$ 
D)  $y ^ { 2 } = 6 x$ 
A)  $y ^ { 2 } = 24 x$ 
B)  $x ^ { 2 } = 6 y$ 
C)  $x ^ { 2 } = 24 y$ 
D)  $y ^ { 2 } = 6 x$

Question 9

Graph the equation.
-$$\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 10

Find an equation for the ellipse described.
-Center at (3, 3); focus at (9, 3); vertex at (11, 3)

Accepted Answer

A)  $\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 1$ 
B)  $\frac { ( x - 3 ) ^ { 2 } } { 64 } + \frac { ( y - 3 ) ^ { 2 } } { 28 } = 1$ 
C)  $\frac { ( x + 3 ) ^ { 2 } } { 64 } + \frac { ( y + 3 ) ^ { 2 } } { 28 } = 1$ 
D)  $\frac { ( x - 3 ) ^ { 2 } } { 121 } + \frac { ( y + 3 ) ^ { 2 } } { 10 } = 2$ 
A)  $\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 1$ 
B)  $\frac { ( x - 3 ) ^ { 2 } } { 64 } + \frac { ( y - 3 ) ^ { 2 } } { 28 } = 1$ 
C)  $\frac { ( x + 3 ) ^ { 2 } } { 64 } + \frac { ( y + 3 ) ^ { 2 } } { 28 } = 1$ 
D)  $\frac { ( x - 3 ) ^ { 2 } } { 121 } + \frac { ( y + 3 ) ^ { 2 } } { 10 } = 2$

Question 11

Write an equation for the graph.
-

Accepted Answer

A)  $\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$ 
B)  $\frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y - 2 ) ^ { 2 } } { 4 } = 1$ 
C)  $\frac { ( x + 1 ) ^ { 2 } } { 16 } + \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$ 
D)  $\frac { ( x + 1 ) ^ { 2 } } { 4 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$ 
A)  $\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$ 
B)  $\frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y - 2 ) ^ { 2 } } { 4 } = 1$ 
C)  $\frac { ( x + 1 ) ^ { 2 } } { 16 } + \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$ 
D)  $\frac { ( x + 1 ) ^ { 2 } } { 4 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$

Question 12

Solve the problem.
-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 27 feet from the center is to be 16 feet above ground?

Accepted Answer

A) 60.72 ft 
B) 30.36 ft 
C) 50.29 ft 
D) 118.13 ft 
A) 60.72 ft 
B) 30.36 ft 
C) 50.29 ft 
D) 118.13 ft

Question 13

Convert the equation to the standard form for a hyperbola by completing the square.
-$$y ^ { 2 } - 4 x ^ { 2 } - 2 y - 8 x - 7 = 0$$

Accepted Answer

A)  $\frac { ( y - 2 ) ^ { 2 } } { 4 } - ( x + 2 ) ^ { 2 } = 1$ 
B)  $( x + 1 ) ^ { 2 } - \frac { ( y - 1 ) ^ { 2 } } { 4 } = 1$ 
C)  $\frac { ( y - 1 ) ^ { 2 } } { 4 } - ( x + 1 ) ^ { 2 } = 1$ 
D)  $\frac { ( x - 1 ) ^ { 2 } } { 4 } - ( y + 1 ) ^ { 2 } = 1$ 
A)  $\frac { ( y - 2 ) ^ { 2 } } { 4 } - ( x + 2 ) ^ { 2 } = 1$ 
B)  $( x + 1 ) ^ { 2 } - \frac { ( y - 1 ) ^ { 2 } } { 4 } = 1$ 
C)  $\frac { ( y - 1 ) ^ { 2 } } { 4 } - ( x + 1 ) ^ { 2 } = 1$ 
D)  $\frac { ( x - 1 ) ^ { 2 } } { 4 } - ( y + 1 ) ^ { 2 } = 1$

Question 14

Solve the problem.
-A bridge is built in the shape of a semielliptical arch. It has a span of 102 feet. The height of the arch 27 feet from the center is to be 11 feet. Find the height of the arch at its center.

Accepted Answer

A) 11.41 ft 
B) 20.78 ft 
C) 12.97 ft 
D) 27.65 ft 
A) 11.41 ft 
B) 20.78 ft 
C) 12.97 ft 
D) 27.65 ft

Question 15

Find an equation for the ellipse described.
-Vertices at (5, -4)and (5, 8); length of minor axis is 6

Accepted Answer

A)  $\frac { ( x - 5 ) ^ { 2 } } { 36 } + \frac { ( y - 2 ) ^ { 2 } } { 9 } = 1$ 
B)  $\frac { ( x + 5 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$ 
C)  $\frac { ( x - 5 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$ 
D)  $\frac { ( x + 5 ) ^ { 2 } } { 9 } - \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$ 
A)  $\frac { ( x - 5 ) ^ { 2 } } { 36 } + \frac { ( y - 2 ) ^ { 2 } } { 9 } = 1$ 
B)  $\frac { ( x + 5 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$ 
C)  $\frac { ( x - 5 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$ 
D)  $\frac { ( x + 5 ) ^ { 2 } } { 9 } - \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$

Question 16

Find an equation for the hyperbola described.
-Vertices at $( - 7,0 )$ and $( 7,0 )$; foci at $( - 11,0 )$ and $( 11,0 )$

Accepted Answer

A)  $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 121 } = 1$ 
B)  $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 72 } = 1$ 
C)  $\frac { x ^ { 2 } } { 72 } - \frac { y ^ { 2 } } { 49 } = 1$ 
D)  $\frac { x ^ { 2 } } { 121 } - \frac { y ^ { 2 } } { 49 } = 1$ 
A)  $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 121 } = 1$ 
B)  $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 72 } = 1$ 
C)  $\frac { x ^ { 2 } } { 72 } - \frac { y ^ { 2 } } { 49 } = 1$ 
D)  $\frac { x ^ { 2 } } { 121 } - \frac { y ^ { 2 } } { 49 } = 1$

Question 17

Solve the problem.
-A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 160 feet and a maximum height of 40 feet. Find the height of the arch at 10 feet from its center.

Accepted Answer

A) 2.5 ft 
B) 39.4 ft 
C) 0.2 ft 
D) 5 ft 
A) 2.5 ft 
B) 39.4 ft 
C) 0.2 ft 
D) 5 ft

Question 18

Find an equation for the hyperbola described.
-Center at $( 6,7 )$; focus at $( - 1,7 )$; vertex at $( 5,7 )$

Accepted Answer

A)  $\frac { ( x - 7 ) ^ { 2 } } { 48 } - ( y - 6 ) ^ { 2 } = 1$ 
B)  $( x - 7 ) ^ { 2 } - \frac { ( y - 6 ) ^ { 2 } } { 48 } = 1$ 
C)  $( x - 6 ) ^ { 2 } - \frac { ( y - 7 ) ^ { 2 } } { 48 } = 1$ 
D)  $\frac { ( x - 6 ) ^ { 2 } } { 48 } - ( y - 7 ) ^ { 2 } = 1$ 
A)  $\frac { ( x - 7 ) ^ { 2 } } { 48 } - ( y - 6 ) ^ { 2 } = 1$ 
B)  $( x - 7 ) ^ { 2 } - \frac { ( y - 6 ) ^ { 2 } } { 48 } = 1$ 
C)  $( x - 6 ) ^ { 2 } - \frac { ( y - 7 ) ^ { 2 } } { 48 } = 1$ 
D)  $\frac { ( x - 6 ) ^ { 2 } } { 48 } - ( y - 7 ) ^ { 2 } = 1$

Question 19

Graph the equation.
-

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 20

Find the center, transverse axis, vertices, and foci of the hyperbola.
-$25 y ^ { 2 } - 36 x ^ { 2 } = 900$

Accepted Answer

A)  center at $( 0,0 )$
transverse axis is $x$-axis
vertices: $( - 5,0 ) , ( 5,0 )$
foci: $( - \sqrt { 61 } , 0 ) , ( \sqrt { 61 } , 0 )$ 
B)  center at $( 0,0 )$
 transverse axis is $\mathrm { y }$-axis
vertices: $( 0 , - 6 ) , ( 0,6 )$
foci: $( - \sqrt { 61 } , 0 ) , ( \sqrt { 61 } , 0 )$ 
C)  center at $( 0,0 )$
transverse axis is $x$-axis
vertices: $( - 6,0 ) , ( 6,0 )$
foci: $( - 5,0 ) , ( 5,0 )$ 
D)  center at $( 0,0 )$
transverse axis is $y$-axis
vertices at $( 0 , - 6 )$ and $( 0,6 )$
foci at $( 0 , - \sqrt { 61 } )$ and $( 0 , \sqrt { 61 } )$ 
A)  center at $( 0,0 )$
transverse axis is $x$-axis
vertices: $( - 5,0 ) , ( 5,0 )$
foci: $( - \sqrt { 61 } , 0 ) , ( \sqrt { 61 } , 0 )$ 
B)  center at $( 0,0 )$
 transverse axis is $\mathrm { y }$-axis
vertices: $( 0 , - 6 ) , ( 0,6 )$
foci: $( - \sqrt { 61 } , 0 ) , ( \sqrt { 61 } , 0 )$ 
C)  center at $( 0,0 )$
transverse axis is $x$-axis
vertices: $( - 6,0 ) , ( 6,0 )$
foci: $( - 5,0 ) , ( 5,0 )$ 
D)  center at $( 0,0 )$
transverse axis is $y$-axis
vertices at $( 0 , - 6 )$ and $( 0,6 )$
foci at $( 0 , - \sqrt { 61 } )$ and $( 0 , \sqrt { 61 } )$

Find an equation for the ellipse described. -Focus at (-2, 0); vertices at (-8, 0)and (8, 0)

Find an equation for the ellipse described. -Foci at (-3, 4)and (-3, -2); length of major axis is 10

Graph the ellipse and locate the foci. - $9 x ^ { 2 } + 4 y ^ { 2 } = 36$ $Graph the ellipse and locate the foci. - 9 x ^ { 2 } + 4 y ^ { 2 } = 36$

Graph the ellipse and locate the foci. - $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$ $Graph the ellipse and locate the foci. - \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$

Graph the equation. - $( x + 1 ) ^ { 2 } = - 8 ( y + 2 )$ $Graph the equation. - ( x + 1 ) ^ { 2 } = - 8 ( y + 2 )$

Find an equation for the ellipse described. -Center (0, 0); major axis horizontal with length 8; length of minor axis is 4

Find an equation for the ellipse described. -Vertices at (-6, 2)and (14, 2); focus at (12, 2)

Find the equation in standard form of the parabola described. -The vertex has coordinates (0, 0), and the focus has coordinates (6, 0).

Graph the equation. - $\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$ $Graph the equation. - \frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$

Find an equation for the ellipse described. -Center at (3, 3); focus at (9, 3); vertex at (11, 3)

Write an equation for the graph. -

Solve the problem. -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 27 feet from the center is to be 16 feet above ground?

Convert the equation to the standard form for a hyperbola by completing the square. - $y ^ { 2 } - 4 x ^ { 2 } - 2 y - 8 x - 7 = 0$

Solve the problem. -A bridge is built in the shape of a semielliptical arch. It has a span of 102 feet. The height of the arch 27 feet from the center is to be 11 feet. Find the height of the arch at its center.

Find an equation for the ellipse described. -Vertices at (5, -4)and (5, 8); length of minor axis is 6

Find an equation for the hyperbola described. -Vertices at $( - 7,0 )$ and $( 7,0 )$ ; foci at $( - 11,0 )$ and $( 11,0 )$

Solve the problem. -A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 160 feet and a maximum height of 40 feet. Find the height of the arch at 10 feet from its center.

Find an equation for the hyperbola described. -Center at $( 6,7 )$ ; focus at $( - 1,7 )$ ; vertex at $( 5,7 )$

Graph the equation. -

Find the center, transverse axis, vertices, and foci of the hyperbola. - $25 y ^ { 2 } - 36 x ^ { 2 } = 900$

Equations, Inequalities, and Applications

The Rectangular Coordinate System, Lines, and Circles

Functions

Polynomial and Rational Functions

Exponential and Logarithmic Functions and Equations

Systems of Equations and Inequalities

Matrices

Sequences and Series; Counting and Probability

Math Exercises: Sets, Intervals, Absolute Value, and Properties

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Exam 6: Conic Sections

Find an equation for the ellipse described. -Focus at (-2, 0); vertices at (-8, 0)and (8, 0)

Find an equation for the ellipse described. -Foci at (-3, 4)and (-3, -2); length of major axis is 10

Graph the ellipse and locate the foci. - 9x2+4y2=369 x ^ { 2 } + 4 y ^ { 2 } = 369x2+4y2=36

Graph the ellipse and locate the foci. - x24+y216=1\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 14x2​+16y2​=1

Graph the equation. - (x+1)2=−8(y+2)( x + 1 ) ^ { 2 } = - 8 ( y + 2 )(x+1)2=−8(y+2)

Find an equation for the ellipse described. -Center (0, 0); major axis horizontal with length 8; length of minor axis is 4

Find an equation for the ellipse described. -Vertices at (-6, 2)and (14, 2); focus at (12, 2)

Find the equation in standard form of the parabola described. -The vertex has coordinates (0, 0), and the focus has coordinates (6, 0).

Graph the equation. - (x+2)216+(y+1)24=1\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 116(x+2)2​+4(y+1)2​=1

Find an equation for the ellipse described. -Center at (3, 3); focus at (9, 3); vertex at (11, 3)

Write an equation for the graph. -

Solve the problem. -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 27 feet from the center is to be 16 feet above ground?

Convert the equation to the standard form for a hyperbola by completing the square. - y2−4x2−2y−8x−7=0y ^ { 2 } - 4 x ^ { 2 } - 2 y - 8 x - 7 = 0y2−4x2−2y−8x−7=0

Solve the problem. -A bridge is built in the shape of a semielliptical arch. It has a span of 102 feet. The height of the arch 27 feet from the center is to be 11 feet. Find the height of the arch at its center.

Find an equation for the ellipse described. -Vertices at (5, -4)and (5, 8); length of minor axis is 6

Find an equation for the hyperbola described. -Vertices at (−7,0)( - 7,0 )(−7,0) and (7,0)( 7,0 )(7,0) ; foci at (−11,0)( - 11,0 )(−11,0) and (11,0)( 11,0 )(11,0)

Solve the problem. -A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 160 feet and a maximum height of 40 feet. Find the height of the arch at 10 feet from its center.

Find an equation for the hyperbola described. -Center at (6,7)( 6,7 )(6,7) ; focus at (−1,7)( - 1,7 )(−1,7) ; vertex at (5,7)( 5,7 )(5,7)

Graph the equation. -

Find the center, transverse axis, vertices, and foci of the hyperbola. - 25y2−36x2=90025 y ^ { 2 } - 36 x ^ { 2 } = 90025y2−36x2=900

Equations, Inequalities, and Applications

The Rectangular Coordinate System, Lines, and Circles

Functions

Polynomial and Rational Functions

Exponential and Logarithmic Functions and Equations

Systems of Equations and Inequalities

Matrices

Sequences and Series; Counting and Probability

Math Exercises: Sets, Intervals, Absolute Value, and Properties

Filters

Graph the ellipse and locate the foci. - $9 x ^ { 2 } + 4 y ^ { 2 } = 36$ $Graph the ellipse and locate the foci. - 9 x ^ { 2 } + 4 y ^ { 2 } = 36$

Graph the ellipse and locate the foci. - $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$ $Graph the ellipse and locate the foci. - \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$

Graph the equation. - $( x + 1 ) ^ { 2 } = - 8 ( y + 2 )$ $Graph the equation. - ( x + 1 ) ^ { 2 } = - 8 ( y + 2 )$

Graph the equation. - $\frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$ $Graph the equation. - \frac { ( x + 2 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$

Convert the equation to the standard form for a hyperbola by completing the square. - $y ^ { 2 } - 4 x ^ { 2 } - 2 y - 8 x - 7 = 0$

Find an equation for the hyperbola described. -Vertices at $( - 7,0 )$ and $( 7,0 )$ ; foci at $( - 11,0 )$ and $( 11,0 )$

Find an equation for the hyperbola described. -Center at $( 6,7 )$ ; focus at $( - 1,7 )$ ; vertex at $( 5,7 )$

Find the center, transverse axis, vertices, and foci of the hyperbola. - $25 y ^ { 2 } - 36 x ^ { 2 } = 900$