Question 1

Find the center, foci, and asymptotes of the hyperbola.
- x2−y2=32x ^ { 2 } - y ^ { 2 } = 32x2−y2=32

Accepted Answer

A)   C:  (2,2) ;F:(8,0) ,(−8,0) ;A:y=2x,y=−2x( 2,2 )  ; F : ( 8,0 )  , ( - 8,0 )  ; A : y = 2 x , y = - 2 x(2,2) ;F:(8,0) ,(−8,0) ;A:y=2x,y=−2x 
B)    C:(0,0) ;F:(4,0) ,(−4,0) ;A:y=x,y=−xC : ( 0,0 )  ; F : ( 4,0 )  , ( - 4,0 )  ; A : y = x , y = - xC:(0,0) ;F:(4,0) ,(−4,0) ;A:y=x,y=−x 
C)   C:  (0,0) ;F:(8,0) ,(−8,0) ;A:y=x,y=−x( 0,0 )  ; F : ( 8,0 )  , ( - 8,0 )  ; A : y = x , y = - x(0,0) ;F:(8,0) ,(−8,0) ;A:y=x,y=−x 
D)   C:  (0,0) ;F:(8,0) ,(−8,0) ;A:y=2x,y=−2x( 0,0 )  ; F : ( 8,0 )  , ( - 8,0 )  ; A : y = 2 x , y = - 2 x(0,0) ;F:(8,0) ,(−8,0) ;A:y=2x,y=−2xE) A) and B)
F) A) and C)

Question 2

Graph the conic section.
- y2−12y+4x+48=0y ^ { 2 } - 12 y + 4 x + 48 = 0y2−12y+4x+48=0

Accepted Answer

A)  
B)  
C)  
D)   E) None of the above
F) A) and B)

Question 3

Find the eccentricity e of the ellipse.
-An elliptical riding path is to be built on a rectangular piece of property that measures 10 mi by 6 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4
Sides

Accepted Answer

A)    x225+y26=1\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 6 } = 125x2​+6y2​=1 
B)    x29+y2100=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 100 } = 19x2​+100y2​=1 
C)    x29+y225=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 19x2​+25y2​=1 
D)    x2100+y29=1\frac { x ^ { 2 } } { 100 } + \frac { y ^ { 2 } } { 9 } = 1100x2​+9y2​=1E) B) and C)
F) A) and D)

Question 4

Find the eccentricity of the conic section shown in the graph.
-If a satellite is scheduled to leave Earth's gravitational influence, its velocity must be increased so that its trajectory changes from elliptical to hyperbolic. Determine the minimum increase in
Velocity necessary for a satellite traveling at a velocity of 5698 meters per second to escape Earth's
Gravitational influence when  D=142×106D = 142 	imes 10 ^ { 6 }D=142×106  m.

Accepted Answer

A)   5773 m per sec
B)   8064 m per sec
C)   5623 m per sec
D)   -3332 m per sec E) A) and D)
F) A) and C)

Question 5

Identify the equation as a parabola, circle, ellipse, or hyperbola.
- 4x2+16y2=644 x ^ { 2 } + 16 y ^ { 2 } = 644x2+16y2=64

Accepted Answer

A)   Hyperbola
B)   Parabola
C)   Ellipse
D)   Circle E) B) and C)
F) A) and B)

Question 6

Write an equation for the hyperbola.
-The roof of a building is in the shape of the hyperbola  y2−2x2=56y ^ { 2 } - 2 x ^ { 2 } = 56y2−2x2=56 , where  xxx  and  yyy  are in meters. Refer to the figure and determine the height,  hhh , of the outside walls.
 A=3 mA = 3 \mathrm {~m}A=3 m

Accepted Answer

A)    3.3 m3.3 \mathrm {~m}3.3 m 
B)    11 m11 \mathrm {~m}11 m 
C)    8.6 m8.6 \mathrm {~m}8.6 m 
D)    74 m74 \mathrm {~m}74 mE) A) and D)
F) B) and D)

Question 7

Graph the parabola.
- x=(y−3) 2x=(y-3) ^{2}x=(y−3) 2

Accepted Answer

A)  
B)  
C)  
D)   E) B) and C)
F) A) and D)

Question 8

Match the equation of the parabola with the appropriate description.
- y+2=2(x+15) 2y + 2 = 2 ( x + 15 )  ^ { 2 }y+2=2(x+15) 2

Accepted Answer

A)   Vertex at  (−15,−2) ( - 15 , - 2 ) (−15,−2)  
B)   Vertex at  (15,−2) ( 15 , - 2 ) (15,−2)  
C)   Vertex at  (2,15) ( 2,15 ) (2,15)  
D)   Vertex at  (−2,15) ( - 2,15 ) (−2,15) E) A) and D)
F) A) and C)

Question 9

Graph the equation. Give the domain and range. Identify whether the equation is a function.
- y9=1−x249\frac { y } { 9 } = \sqrt { 1 - \frac { x ^ { 2 } } { 49 } }9y​=1−49x2​​

Accepted Answer

A)   [−7,7];[0,1]; not a function [-7,7] ;[0,1] ; 	ext { not a function }[−7,7];[0,1]; not a function  
B)   [−7,7];[0,9]; function [-7,7] ;[0,9] ; 	ext { function }[−7,7];[0,9]; function  
C)   [−7,7];[−1,0];[ - 7,7 ] ; [ - 1,0 ] ;[−7,7];[−1,0];  not a function
D)   [−7,7];[−9,0][ - 7,7 ] ; [ - 9,0 ][−7,7];[−9,0] ; function  E) A) and B)
F) A) and C)

Question 10

Provide an appropriate response.
-The eccentricity of ellipse $\mathrm { A }$ is $\mathrm { e } _ { \mathrm { A } } = 0.18$ and the eccentricity of ellipse $\mathrm { B }$ is $\mathrm { eB } = 0.82$. Which ellipse is flatter?

Accepted Answer

The answer of Provide an appropriate response.
-The eccentricity of ellipse...

Question 11

Provide an appropriate response.
-Suppose that $A C > 0$ and $A \neq C$ in the equation $A x ^ { 2 } + B x + C y ^ { 2 } + D y + E = 0$. What kind of graph does this equation have?

Accepted Answer

The equati...

Question 12

Graph the conic section.
- 9x2−16y2−36x−96y−252=09 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 09x2−16y2−36x−96y−252=0

Accepted Answer

A)  
B)  
C)  
D)   E) A) and D)
F) B) and C)

Question 13

Write an equation for the parabola.
-vertex:  (−4,3) ( - 4,3 ) (−4,3)  , directrix:  x=−6x = - 6x=−6

Accepted Answer

A)    (y−3) 2=8(x+4) ( y - 3 )  ^ { 2 } = 8 ( x + 4 ) (y−3) 2=8(x+4)  
B)    (x−3) 2=8(y+4) ( x - 3 )  ^ { 2 } = 8 ( y + 4 ) (x−3) 2=8(y+4)  
C)    (x+4) 2=−8(x−3) ( x + 4 )  ^ { 2 } = - 8 ( x - 3 ) (x+4) 2=−8(x−3)  
D)    (y−3) 2=−8(x+4) ( y - 3 )  ^ { 2 } = - 8 ( x + 4 ) (y−3) 2=−8(x+4) E) None of the above
F) A) and B)

Question 14

Match the equation of the parabola with the appropriate description.
- x+2=2(y+7) 2x + 2 = 2 ( y + 7 )  ^ { 2 }x+2=2(y+7) 2

Accepted Answer

A)   Vertex at  (−7,−2) ( - 7 , - 2 ) (−7,−2)  
B)   Vertex at  (2,7) ( 2,7 ) (2,7)  
C)   Vertex at  (−2,−7) ( - 2 , - 7 ) (−2,−7)  
D)   Vertex at  (7,2) ( 7,2 ) (7,2) E) B) and D)
F) A) and D)

Question 15

Write an equation for the hyperbola.
-A nuclear cooling tower cross section is a hyperbola having a diameter of  60ft60 \mathrm { ft }60ft  at the center. The distance between the two foci is  92ft92 \mathrm { ft }92ft . What is the equation for the hyperbola?

Accepted Answer

A)    x2900−y21216=5\frac { x ^ { 2 } } { 900 } - \frac { y ^ { 2 } } { 1216 } = 5900x2​−1216y2​=5 
B)    x2900−y21218=1\frac { x ^ { 2 } } { 900 } - \frac { y ^ { 2 } } { 1218 } = 1900x2​−1218y2​=1 
C)    x2900−y22116=1\frac { x ^ { 2 } } { 900 } - \frac { y ^ { 2 } } { 2116 } = 1900x2​−2116y2​=1 
D)    x2900−y21216=1\frac { x ^ { 2 } } { 900 } - \frac { y ^ { 2 } } { 1216 } = 1900x2​−1216y2​=1E) A) and D)
F) A) and C)

Question 16

Determine the two equations necessary to graph the ellipse with a graphing calculator.
- x264+(y−2) 281=1\frac { x ^ { 2 } } { 64 } + \frac { ( y - 2 )  ^ { 2 } } { 81 } = 164x2​+81(y−2) 2​=1

Accepted Answer

A)    y=95184−4x216y = \frac { 9 \sqrt { 5184 - 4 x ^ { 2 } } } { 16 }y=1695184−4x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​  and  y=−95184−4x216y = - \frac { 9 \sqrt { 5184 - 4 x ^ { 2 } } } { 16 }y=−1695184−4x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
B)    y=16−964−x28y = \frac { 16 - 9 \sqrt { 64 - x ^ { 2 } } } { 8 }y=816−964−x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​  and  y=964−x2+168y = \frac { 9 \sqrt { 64 - x ^ { 2 } } + 16 } { 8 }y=8964−x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+16​ 
C)    y=5440−81x28y = \frac { \sqrt { 5440 - 81 x ^ { 2 } } } { 8 }y=85440−81x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​  and  y=−5440−81x28y = - \frac { \sqrt { 5440 - 81 x ^ { 2 } } } { 8 }y=−85440−81x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    y=256−81x28y = \frac { \sqrt { 256 - 81 x ^ { 2 } } } { 8 }y=8256−81x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​  and  y=−256−81x28y = - \frac { \sqrt { 256 - 81 x ^ { 2 } } } { 8 }y=−8256−81x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​E) A) and C)
F) B) and D)

Question 17

Write an equation for the hyperbola.
-vertices at  (0,3) ,(0,−3) ( 0,3 )  , ( 0 , - 3 ) (0,3) ,(0,−3)  ; foci at  (0,7) ,(0,−7) ( 0,7 )  , ( 0 , - 7 ) (0,7) ,(0,−7)

Accepted Answer

A)    y249−x29=1\frac { y ^ { 2 } } { 49 } - \frac { x ^ { 2 } } { 9 } = 149y2​−9x2​=1 
B)    y240−x29=1\frac { y ^ { 2 } } { 40 } - \frac { x ^ { 2 } } { 9 } = 140y2​−9x2​=1 
C)    y29−x249=1\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 49 } = 19y2​−49x2​=1 
D)    y29−x240=1\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 19y2​−40x2​=1E) B) and C)
F) None of the above

Question 18

Write an equation for the ellipse.
-foci at  (±42,0) ;x( \pm 4 \sqrt { 2 } , 0 )  ; x(±42​,0) ;x -intercepts  (±9,0) ( \pm 9,0 ) (±9,0)

Accepted Answer

A)    x281+y249=1\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 49 } = 181x2​+49y2​=1 
B)    x29+y27=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 7 } = 19x2​+7y2​=1 
C)    x27+y29=1\frac { x ^ { 2 } } { 7 } + \frac { y ^ { 2 } } { 9 } = 17x2​+9y2​=1 
D)    x249+y281=1\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 81 } = 149x2​+81y2​=1E) A) and B)
F) A) and C)

Question 19

Find the eccentricity of the conic section shown in the graph.
-X = - 1/2

Accepted Answer

A)    e=3e = 3e=3 
B)    e=1e = 1e=1 
C)    e=2e = 2e=2 
D)    e=12e = \frac { 1 } { 2 }e=21​E) A) and C)
F) C) and D)

Question 20

Match the equation of the parabola with the appropriate description.
- y=6x2−6x+4y = 6 x ^ { 2 } - 6 x + 4y=6x2−6x+4

Accepted Answer

A)   Opens left
B)   Opens up
C)   Opens down
D)   Opens right E) A) and D)
F) None of the above

Exam 11: Analytic Geometry

Match the equation of the parabola with the appropriate description. - $y + 2 = 2 ( x + 15 ) ^ { 2 }$

Correct Answer
verified

Graph the conic section. - $y ^ { 2 } - 12 y + 4 x + 48 = 0$ $Graph the conic section. - y ^ { 2 } - 12 y + 4 x + 48 = 0 A) B) C) D)$

Correct Answer
verified

Match the equation of the parabola with the appropriate description. - $x + 2 = 2 ( y + 7 ) ^ { 2 }$

Correct Answer
verified

Graph the parabola. - $x=(y-3) ^{2}$ $Graph the parabola. - x=(y-3) ^{2} A) B) C) D)$

Correct Answer
verified

Write an equation for the hyperbola. -vertices at $( 0,3 ) , ( 0 , - 3 )$ ; foci at $( 0,7 ) , ( 0 , - 7 )$

Correct Answer
verified

Write an equation for the ellipse. -foci at $( \pm 4 \sqrt { 2 } , 0 ) ; x$ -intercepts $( \pm 9,0 )$

Correct Answer
verified

Provide an appropriate response. -Suppose that $A C > 0$ and $A \neq C$ in the equation $A x ^ { 2 } + B x + C y ^ { 2 } + D y + E = 0$ . What kind of graph does this equation have?

Correct Answer
verified
The equati...
View Answer

Determine the two equations necessary to graph the ellipse with a graphing calculator. - $\frac { x ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 81 } = 1$

Correct Answer
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Find the center, foci, and asymptotes of the hyperbola. - $x ^ { 2 } - y ^ { 2 } = 32$

Correct Answer
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Write an equation for the parabola. -vertex: $( - 4,3 )$ , directrix: $x = - 6$

Correct Answer
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Graph the conic section. - $9 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 0$ $Graph the conic section. - 9 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 0 A) B) C) D)$

Correct Answer
verified

Find the eccentricity of the conic section shown in the graph. -X = - 1/2 $Find the eccentricity of the conic section shown in the graph. -X = - 1/2 A) e = 3 B) e = 1 C) e = 2 D) e = \frac { 1 } { 2 }$

Correct Answer
verified

Write an equation for the hyperbola. -A nuclear cooling tower cross section is a hyperbola having a diameter of $60 \mathrm { ft }$ at the center. The distance between the two foci is $92 \mathrm { ft }$ . What is the equation for the hyperbola?

Correct Answer
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Correct Answer
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Identify the equation as a parabola, circle, ellipse, or hyperbola. - $4 x ^ { 2 } + 16 y ^ { 2 } = 64$

Correct Answer
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Correct Answer
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Correct Answer
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Correct Answer
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Match the equation of the parabola with the appropriate description. - $y = 6 x ^ { 2 } - 6 x + 4$

Correct Answer
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Provide an appropriate response. -The eccentricity of ellipse $\mathrm { A }$ is $\mathrm { e } _ { \mathrm { A } } = 0.18$ and the eccentricity of ellipse $\mathrm { B }$ is $\mathrm { eB } = 0.82$ . Which ellipse is flatter?

Correct Answer
verified

Exam 11: Analytic Geometry

Match the equation of the parabola with the appropriate description. - y+2=2(x+15) 2y + 2 = 2 ( x + 15 ) ^ { 2 }y+2=2(x+15) 2

Correct AnswerverifiedShow Answer

Graph the conic section. - y2−12y+4x+48=0y ^ { 2 } - 12 y + 4 x + 48 = 0y2−12y+4x+48=0

Correct AnswerverifiedShow Answer

Match the equation of the parabola with the appropriate description. - x+2=2(y+7) 2x + 2 = 2 ( y + 7 ) ^ { 2 }x+2=2(y+7) 2

Correct AnswerverifiedShow Answer

Graph the parabola. - x=(y−3) 2x=(y-3) ^{2}x=(y−3) 2

Correct AnswerverifiedShow Answer

Write an equation for the hyperbola. -vertices at (0,3) ,(0,−3) ( 0,3 ) , ( 0 , - 3 ) (0,3) ,(0,−3) ; foci at (0,7) ,(0,−7) ( 0,7 ) , ( 0 , - 7 ) (0,7) ,(0,−7)

Correct AnswerverifiedShow Answer

Write an equation for the ellipse. -foci at (±42,0) ;x( \pm 4 \sqrt { 2 } , 0 ) ; x(±42​,0) ;x -intercepts (±9,0) ( \pm 9,0 ) (±9,0)

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Provide an appropriate response. -Suppose that AC>0A C > 0AC>0 and A≠CA \neq CA=C in the equation Ax2+Bx+Cy2+Dy+E=0A x ^ { 2 } + B x + C y ^ { 2 } + D y + E = 0Ax2+Bx+Cy2+Dy+E=0 . What kind of graph does this equation have?

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Determine the two equations necessary to graph the ellipse with a graphing calculator. - x264+(y−2) 281=1\frac { x ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 81 } = 164x2​+81(y−2) 2​=1

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Find the center, foci, and asymptotes of the hyperbola. - x2−y2=32x ^ { 2 } - y ^ { 2 } = 32x2−y2=32

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Write an equation for the parabola. -vertex: (−4,3) ( - 4,3 ) (−4,3) , directrix: x=−6x = - 6x=−6

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Graph the conic section. - 9x2−16y2−36x−96y−252=09 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 09x2−16y2−36x−96y−252=0

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Find the eccentricity of the conic section shown in the graph. -X = - 1/2

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Write an equation for the hyperbola. -A nuclear cooling tower cross section is a hyperbola having a diameter of 60ft60 \mathrm { ft }60ft at the center. The distance between the two foci is 92ft92 \mathrm { ft }92ft . What is the equation for the hyperbola?

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Graph the equation. Give the domain and range. Identify whether the equation is a function. - y9=1−x249\frac { y } { 9 } = \sqrt { 1 - \frac { x ^ { 2 } } { 49 } }9y​=1−49x2​​

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Identify the equation as a parabola, circle, ellipse, or hyperbola. - 4x2+16y2=644 x ^ { 2 } + 16 y ^ { 2 } = 644x2+16y2=64

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Write an equation for the hyperbola. -The roof of a building is in the shape of the hyperbola y2−2x2=56y ^ { 2 } - 2 x ^ { 2 } = 56y2−2x2=56 , where xxx and yyy are in meters. Refer to the figure and determine the height, hhh , of the outside walls. A=3 mA = 3 \mathrm {~m}A=3 m

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Find the eccentricity e of the ellipse. -An elliptical riding path is to be built on a rectangular piece of property that measures 10 mi by 6 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 Sides

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Match the equation of the parabola with the appropriate description. - y=6x2−6x+4y = 6 x ^ { 2 } - 6 x + 4y=6x2−6x+4

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Provide an appropriate response. -The eccentricity of ellipse A\mathrm { A }A is eA=0.18\mathrm { e } _ { \mathrm { A } } = 0.18eA​=0.18 and the eccentricity of ellipse B\mathrm { B }B is eB=0.82\mathrm { eB } = 0.82eB=0.82 . Which ellipse is flatter?

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Match the equation of the parabola with the appropriate description. - $y + 2 = 2 ( x + 15 ) ^ { 2 }$

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Graph the conic section. - $y ^ { 2 } - 12 y + 4 x + 48 = 0$ $Graph the conic section. - y ^ { 2 } - 12 y + 4 x + 48 = 0 A) B) C) D)$

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Match the equation of the parabola with the appropriate description. - $x + 2 = 2 ( y + 7 ) ^ { 2 }$

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Graph the parabola. - $x=(y-3) ^{2}$ $Graph the parabola. - x=(y-3) ^{2} A) B) C) D)$

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Write an equation for the hyperbola. -vertices at $( 0,3 ) , ( 0 , - 3 )$ ; foci at $( 0,7 ) , ( 0 , - 7 )$

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Write an equation for the ellipse. -foci at $( \pm 4 \sqrt { 2 } , 0 ) ; x$ -intercepts $( \pm 9,0 )$

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Provide an appropriate response. -Suppose that $A C > 0$ and $A \neq C$ in the equation $A x ^ { 2 } + B x + C y ^ { 2 } + D y + E = 0$ . What kind of graph does this equation have?

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Determine the two equations necessary to graph the ellipse with a graphing calculator. - $\frac { x ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 81 } = 1$

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Find the center, foci, and asymptotes of the hyperbola. - $x ^ { 2 } - y ^ { 2 } = 32$

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Write an equation for the parabola. -vertex: $( - 4,3 )$ , directrix: $x = - 6$

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Graph the conic section. - $9 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 0$ $Graph the conic section. - 9 x ^ { 2 } - 16 y ^ { 2 } - 36 x - 96 y - 252 = 0 A) B) C) D)$

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Find the eccentricity of the conic section shown in the graph. -X = - 1/2 $Find the eccentricity of the conic section shown in the graph. -X = - 1/2 A) e = 3 B) e = 1 C) e = 2 D) e = \frac { 1 } { 2 }$

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Write an equation for the hyperbola. -A nuclear cooling tower cross section is a hyperbola having a diameter of $60 \mathrm { ft }$ at the center. The distance between the two foci is $92 \mathrm { ft }$ . What is the equation for the hyperbola?

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Identify the equation as a parabola, circle, ellipse, or hyperbola. - $4 x ^ { 2 } + 16 y ^ { 2 } = 64$

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Match the equation of the parabola with the appropriate description. - $y = 6 x ^ { 2 } - 6 x + 4$

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Provide an appropriate response. -The eccentricity of ellipse $\mathrm { A }$ is $\mathrm { e } _ { \mathrm { A } } = 0.18$ and the eccentricity of ellipse $\mathrm { B }$ is $\mathrm { eB } = 0.82$ . Which ellipse is flatter?

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