Question 1

Name the conic.
-

Accepted Answer

A)  hyperbola 
B)  ellipse 
C)  circle 
D)  parabola 
A)  hyperbola 
B)  ellipse 
C)  circle 
D)  parabola B

Question 2

Find the center, foci, and vertices of the ellipse.
-$$\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$$

Accepted Answer

A)  center at $( 3 , - 3 )$
foci at $( 3 + 3 \sqrt { 3 } , - 3 ) , ( 3 - 3 \sqrt { 3 } , - 3 )$
vertices at $( - 3 , - 3 ) , ( 9 , - 3 )$ 
B)  center at $( 3 , - 3 )$
foci at $( - 3 \sqrt { 3 } , - 3 ) , ( 3 \sqrt { 3 } , - 3 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$ 
C)  center at $( - 3,3 )$
foci at $( - 3 + 3 \sqrt { 3 } , 3 ) , ( - 3 - 3 \sqrt { 3 } , 3 )$
vertices at $( - 3 , - 3 ) , ( 9 , - 3 )$ 
D)  center at $( 3 , - 3 )$
foci at $( 3 + 3 \sqrt { 3 } , 3 ) , ( 3 - 3 \sqrt { 3 } , 3 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$ 
A)  center at $( 3 , - 3 )$
foci at $( 3 + 3 \sqrt { 3 } , - 3 ) , ( 3 - 3 \sqrt { 3 } , - 3 )$
vertices at $( - 3 , - 3 ) , ( 9 , - 3 )$ 
B)  center at $( 3 , - 3 )$
foci at $( - 3 \sqrt { 3 } , - 3 ) , ( 3 \sqrt { 3 } , - 3 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$ 
C)  center at $( - 3,3 )$
foci at $( - 3 + 3 \sqrt { 3 } , 3 ) , ( - 3 - 3 \sqrt { 3 } , 3 )$
vertices at $( - 3 , - 3 ) , ( 9 , - 3 )$ 
D)  center at $( 3 , - 3 )$
foci at $( 3 + 3 \sqrt { 3 } , 3 ) , ( 3 - 3 \sqrt { 3 } , 3 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$ A

Question 3

Graph the ellipse and locate the foci.
-$4 x^{2}+16 y^{2}=64$

Accepted Answer

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

Question 4

Find the asymptotes of the hyperbola.
-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 = 125 x 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).

Accepted Answer

A)  $\frac { y ^ { 2 } } { \frac { 25 } { 9 } } - \frac { x ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { \left( \frac { 36 } { 5 } \right) ^ { 2 } } = 1$ 
C)  $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { \frac { 25 } { 9 } } = 1$ 
D)  $\frac { x ^ { 2 } } { \left( \frac { 36 } { 5 } \right) ^ { 2 } } - \frac { y ^ { 2 } } { 16 } = 1$ 
A)  $\frac { y ^ { 2 } } { \frac { 25 } { 9 } } - \frac { x ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { \left( \frac { 36 } { 5 } \right) ^ { 2 } } = 1$ 
C)  $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { \frac { 25 } { 9 } } = 1$ 
D)  $\frac { x ^ { 2 } } { \left( \frac { 36 } { 5 } \right) ^ { 2 } } - \frac { y ^ { 2 } } { 16 } = 1$

Question 5

Find the center, foci, and vertices of the ellipse.
-$2 x ^ { 2 } + 5 y ^ { 2 } - 20 x + 30 y + 85 = 0$

Accepted Answer

A)  $\frac { ( x - 5 ) ^ { 2 } } { 2 } + \frac { ( y + 3 ) ^ { 2 } } { 5 } = 1$
center: $( 5 , - 3 )$; foci: $( 6.7 , - 3 ) , ( 3.3 , - 3 )$; vertices: $( 7.2 , - 3 ) , ( 2.8 , - 3 )$ 
B)  $\frac { ( x - 5 ) ^ { 2 } } { 5 } + \frac { ( y + 3 ) ^ { 2 } } { 2 } = 1$
center: $( - 5,3 )$; foci: $( - 3.3,3 ) , ( - 6.7,3 )$; vertices: $( - 7.2,3 ) , ( - 2.8,3 )$ 
C)  $\frac { ( x - 5 ) ^ { 2 } } { 2 } + \frac { ( y + 3 ) ^ { 2 } } { 5 } = 1$
center: $( - 5,3 )$; foci: $( - 3.3,3 ) , ( - 6.7,3 )$; vertices: $( - 7.2,3 ) , ( - 2.8,3 )$ 
D)  $\frac { ( x - 5 ) ^ { 2 } } { 5 } + \frac { ( y + 3 ) ^ { 2 } } { 2 } = 1$
center: $( 5 , - 3 )$; foci: $( 6.7 , - 3 ) , ( 3.3 , - 3 )$; vertices: $( 7.2 , - 3 ) , ( 2.8 , - 3 )$ 
A)  $\frac { ( x - 5 ) ^ { 2 } } { 2 } + \frac { ( y + 3 ) ^ { 2 } } { 5 } = 1$
center: $( 5 , - 3 )$; foci: $( 6.7 , - 3 ) , ( 3.3 , - 3 )$; vertices: $( 7.2 , - 3 ) , ( 2.8 , - 3 )$ 
B)  $\frac { ( x - 5 ) ^ { 2 } } { 5 } + \frac { ( y + 3 ) ^ { 2 } } { 2 } = 1$
center: $( - 5,3 )$; foci: $( - 3.3,3 ) , ( - 6.7,3 )$; vertices: $( - 7.2,3 ) , ( - 2.8,3 )$ 
C)  $\frac { ( x - 5 ) ^ { 2 } } { 2 } + \frac { ( y + 3 ) ^ { 2 } } { 5 } = 1$
center: $( - 5,3 )$; foci: $( - 3.3,3 ) , ( - 6.7,3 )$; vertices: $( - 7.2,3 ) , ( - 2.8,3 )$ 
D)  $\frac { ( x - 5 ) ^ { 2 } } { 5 } + \frac { ( y + 3 ) ^ { 2 } } { 2 } = 1$
center: $( 5 , - 3 )$; foci: $( 6.7 , - 3 ) , ( 3.3 , - 3 )$; vertices: $( 7.2 , - 3 ) , ( 2.8 , - 3 )$

Question 6

Find an equation for the ellipse described.
-Center at $( 4,5 )$; focus at $( 9,5 )$; vertex at $( 11,5 )$

Accepted Answer

A)  $\frac { ( x - 4 ) ^ { 2 } } { 100 } + \frac { ( y + 5 ) ^ { 2 } } { 9 } = 2$ 
B)  $\frac { ( x + 4 ) ^ { 2 } } { 49 } + \frac { ( y + 5 ) ^ { 2 } } { 24 } = 1$ 
C)  $\frac { ( x + 4 ) ^ { 2 } } { 25 } - \frac { ( y - 5 ) ^ { 2 } } { 19 } = 1$ 
D)  $\frac { ( x - 4 ) ^ { 2 } } { 49 } + \frac { ( y - 5 ) ^ { 2 } } { 24 } = 1$ 
A)  $\frac { ( x - 4 ) ^ { 2 } } { 100 } + \frac { ( y + 5 ) ^ { 2 } } { 9 } = 2$ 
B)  $\frac { ( x + 4 ) ^ { 2 } } { 49 } + \frac { ( y + 5 ) ^ { 2 } } { 24 } = 1$ 
C)  $\frac { ( x + 4 ) ^ { 2 } } { 25 } - \frac { ( y - 5 ) ^ { 2 } } { 19 } = 1$ 
D)  $\frac { ( x - 4 ) ^ { 2 } } { 49 } + \frac { ( y - 5 ) ^ { 2 } } { 24 } = 1$

Question 7

Solve the problem.
-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?

Accepted Answer

A)  41.6 ft 
B)  17.7 ft 
C)  35.5 ft 
D)  20.8 ft 
A)  41.6 ft 
B)  17.7 ft 
C)  35.5 ft 
D)  20.8 ft

Question 8

Solve the problem.
-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?

Accepted Answer

The answer of Solve the problem.
-A sealed-beam headlight is in...

Question 9

Name the conic.
-

Accepted Answer

A)  parabola 
B)  ellipse 
C)  circle 
D)  hyperbola 
A)  parabola 
B)  ellipse 
C)  circle 
D)  hyperbola

Question 10

Find an equation for the ellipse described. Graph the equation.
-Foci at (0, 3) and (6, 3); length of major axis is 10

Accepted Answer

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

Question 11

Solve the problem.
-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. At some point along the road from the lowest
Point of the cable, the cable is 0.81 inches above the roadway. Find the distance between that point and the base
Of the nearest tower.

Accepted Answer

A)  21.2 in. 
B)  9.2 in. 
C)  21 in. 
D)  8.8 in. 
A)  21.2 in. 
B)  9.2 in. 
C)  21 in. 
D)  8.8 in.

Question 12

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 25 feet from the center is to be 11 feet. Find the height of the arch at its center.

Accepted Answer

A)  25.6 ft 
B)  22.44 ft 
C)  12.62 ft 
D)  11.35 ft 
A)  25.6 ft 
B)  22.44 ft 
C)  12.62 ft 
D)  11.35 ft

Question 13

Match the equation to the graph.
-$16 x^{2}=9 y^{2}+144$

Accepted Answer

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

Question 14

Match the equation to the graph.
-$$\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 4 } = 1$$

Accepted Answer

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

Question 15

Solve the problem.
-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 15 feet above ground?

Accepted Answer

A)  117.17 ft 
B)  116 ft 
C)  33.49 ft 
D)  66.97 ft 
A)  117.17 ft 
B)  116 ft 
C)  33.49 ft 
D)  66.97 ft

Question 16

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

Accepted Answer

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

Question 17

Find an equation for the ellipse described. Graph the equation.
-Center at (-2, 4); focus at (-8, 4); contains the point (-10, 4)

Accepted Answer

A)  $\frac { ( x + 4 ) ^ { 2 } } { 28 } + \frac { ( y - 2 ) ^ { 2 } } { 64 } = 1$
  
B)  $\frac { ( x + 4 ) ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 28 } = 1$
  
C)  $\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 28 } = 1$
  
D)  $\frac { ( x + 2 ) ^ { 2 } } { 28 } + \frac { ( y - 4 ) ^ { 2 } } { 64 } = 1$
  
A)  $\frac { ( x + 4 ) ^ { 2 } } { 28 } + \frac { ( y - 2 ) ^ { 2 } } { 64 } = 1$
  
B)  $\frac { ( x + 4 ) ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 28 } = 1$
  
C)  $\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 28 } = 1$
  
D)  $\frac { ( x + 2 ) ^ { 2 } } { 28 } + \frac { ( y - 4 ) ^ { 2 } } { 64 } = 1$

Question 18

Find the center, foci, and vertices of the ellipse.
-$$16 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 64$$

Accepted Answer

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

Question 19

Write an equation for the parabola.
-

Accepted Answer

A)  y2 = 4x 
B)  y2 = -4x 
C)  x2 = -4y 
D)  x2 = 4y 
A)  y2 = 4x 
B)  y2 = -4x 
C)  x2 = -4y 
D)  x2 = 4y

Question 20

Find the center, foci, and vertices of the ellipse.
-$$25 x ^ { 2 } + y ^ { 2 } - 450 x + 2000 = 0$$

Accepted Answer

A)  $( x - 9 ) ^ { 2 } + \frac { y ^ { 2 } } { 25 } = 1$
center: $( 9,0 )$; foci: $( 9,2 \sqrt { 6 } ) , ( 9 , - 2 \sqrt { 6 } )$; vertices: $( 9,5 ) , ( 9 , - 5 )$ 
B)  $\frac { x ^ { 2 } } { 25 } + ( y - 9 ) ^ { 2 } = 1$
center: $( 9,0 )$; foci: $( 9,2 \sqrt { 6 } ) , ( 9 , - 2 \sqrt { 6 } )$; vertices: $( 9,5 ) , ( 9 , - 5 )$ 
C)  $\frac { x ^ { 2 } } { 81 } + ( y - 5 ) ^ { 2 } = 1$
center: $( 5,0 )$; foci: $( 5,4 \sqrt { 5 } ) , ( 5 , - 2 \sqrt { 6 } )$; vertices: $( 5,9 )$, $( 5 , - 9 )$ 
D)  $( x - 5 ) ^ { 2 } + \frac { y ^ { 2 } } { 81 } = 1$
center: $( 5,0 )$; foci: $( 5,4 \sqrt { 5 } ) , ( 5 , - 4 \sqrt { 5 } )$; vertices: $( 5,9 ) , ( 5 , - 9 )$ 
A)  $( x - 9 ) ^ { 2 } + \frac { y ^ { 2 } } { 25 } = 1$
center: $( 9,0 )$; foci: $( 9,2 \sqrt { 6 } ) , ( 9 , - 2 \sqrt { 6 } )$; vertices: $( 9,5 ) , ( 9 , - 5 )$ 
B)  $\frac { x ^ { 2 } } { 25 } + ( y - 9 ) ^ { 2 } = 1$
center: $( 9,0 )$; foci: $( 9,2 \sqrt { 6 } ) , ( 9 , - 2 \sqrt { 6 } )$; vertices: $( 9,5 ) , ( 9 , - 5 )$ 
C)  $\frac { x ^ { 2 } } { 81 } + ( y - 5 ) ^ { 2 } = 1$
center: $( 5,0 )$; foci: $( 5,4 \sqrt { 5 } ) , ( 5 , - 2 \sqrt { 6 } )$; vertices: $( 5,9 )$, $( 5 , - 9 )$ 
D)  $( x - 5 ) ^ { 2 } + \frac { y ^ { 2 } } { 81 } = 1$
center: $( 5,0 )$; foci: $( 5,4 \sqrt { 5 } ) , ( 5 , - 4 \sqrt { 5 } )$; vertices: $( 5,9 ) , ( 5 , - 9 )$

Name the conic. -

Find the center, foci, and vertices of the ellipse. - $\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$

Graph the ellipse and locate the foci. - $4 x^{2}+16 y^{2}=64$

Find the center, foci, and vertices of the ellipse. - $2 x ^ { 2 } + 5 y ^ { 2 } - 20 x + 30 y + 85 = 0$

Find an equation for the ellipse described. -Center at $( 4,5 )$ ; focus at $( 9,5 )$ ; vertex at $( 11,5 )$

Solve the problem. -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?

Solve the problem. -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?

Name the conic. -

Find an equation for the ellipse described. Graph the equation. -Foci at (0, 3) and (6, 3); length of major axis is 10

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 25 feet from the center is to be 11 feet. Find the height of the arch at its center.

Match the equation to the graph. - $16 x^{2}=9 y^{2}+144$

Match the equation to the graph. - $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 4 } = 1$

Solve the problem. -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 15 feet above ground?

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

Find an equation for the ellipse described. Graph the equation. -Center at (-2, 4); focus at (-8, 4); contains the point (-10, 4)

Find the center, foci, and vertices of the ellipse. - $16 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 64$ $Find the center, foci, and vertices of the ellipse. - 16 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 64$

Write an equation for the parabola. -

Find the center, foci, and vertices of the ellipse. - $25 x ^ { 2 } + y ^ { 2 } - 450 x + 2000 = 0$

Functions and Their Graphs

Linear and Quadratic Functions

Polynomial and Rational Functions

Systems of Equations and Inequalities

Sequences; Induction; the Binomial Theorem

Counting and Probability

Filters

Exam 5: Conics

Name the conic. -

Find the center, foci, and vertices of the ellipse. - (x−3)236+(y+3)29=1\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 136(x−3)2​+9(y+3)2​=1

Graph the ellipse and locate the foci. - 4x2+16y2=644 x^{2}+16 y^{2}=644x2+16y2=64

Find the center, foci, and vertices of the ellipse. - 2x2+5y2−20x+30y+85=02 x ^ { 2 } + 5 y ^ { 2 } - 20 x + 30 y + 85 = 02x2+5y2−20x+30y+85=0

Find an equation for the ellipse described. -Center at (4,5)( 4,5 )(4,5) ; focus at (9,5)( 9,5 )(9,5) ; vertex at (11,5)( 11,5 )(11,5)

Solve the problem. -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?

Solve the problem. -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?

Name the conic. -

Find an equation for the ellipse described. Graph the equation. -Foci at (0, 3) and (6, 3); length of major axis is 10

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 25 feet from the center is to be 11 feet. Find the height of the arch at its center.

Match the equation to the graph. - 16x2=9y2+14416 x^{2}=9 y^{2}+14416x2=9y2+144

Match the equation to the graph. - y216−x24=1\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 4 } = 116y2​−4x2​=1

Solve the problem. -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 15 feet above ground?

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

Find an equation for the ellipse described. Graph the equation. -Center at (-2, 4); focus at (-8, 4); contains the point (-10, 4)

Find the center, foci, and vertices of the ellipse. - 16(x−1)2+4(y+2)2=6416 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 6416(x−1)2+4(y+2)2=64

Write an equation for the parabola. -

Find the center, foci, and vertices of the ellipse. - 25x2+y2−450x+2000=025 x ^ { 2 } + y ^ { 2 } - 450 x + 2000 = 025x2+y2−450x+2000=0

Functions and Their Graphs

Linear and Quadratic Functions

Polynomial and Rational Functions

Systems of Equations and Inequalities

Sequences; Induction; the Binomial Theorem

Counting and Probability

Filters

Find the center, foci, and vertices of the ellipse. - $\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$

Graph the ellipse and locate the foci. - $4 x^{2}+16 y^{2}=64$

Find the center, foci, and vertices of the ellipse. - $2 x ^ { 2 } + 5 y ^ { 2 } - 20 x + 30 y + 85 = 0$

Find an equation for the ellipse described. -Center at $( 4,5 )$ ; focus at $( 9,5 )$ ; vertex at $( 11,5 )$

Match the equation to the graph. - $16 x^{2}=9 y^{2}+144$

Match the equation to the graph. - $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 4 } = 1$

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

Find the center, foci, and vertices of the ellipse. - $16 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 64$ $Find the center, foci, and vertices of the ellipse. - 16 ( x - 1 ) ^ { 2 } + 4 ( y + 2 ) ^ { 2 } = 64$

Find the center, foci, and vertices of the ellipse. - $25 x ^ { 2 } + y ^ { 2 } - 450 x + 2000 = 0$