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Identify the equation without completing the square. - 2y2−2x−3y=02 y ^ { 2 } - 2 x - 3 y = 0


A) hyperbola
B) ellipse
C) parabola
D) not a conic

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Discuss the equation and graph it. - r=42−sin⁡θr = \frac { 4 } { 2 - \sin \theta }


A) directrix perpendicular to polar axis 8 right of pole
center (−815,0) \left(-\frac{8}{15}, 0\right)
vertices (83,π) ,(85,0) \left(\frac{8}{3}, \pi\right) ,\left(\frac{8}{5}, 0\right)
Discuss the equation and graph it. - r = \frac { 4 } { 2 - \sin \theta } A) directrix perpendicular to polar axis 8 right of pole center \left(-\frac{8}{15}, 0\right) vertices \left(\frac{8}{3}, \pi\right) ,\left(\frac{8}{5}, 0\right) B) directrix parallel to polar axis 8 above pole center \left(-\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(-\frac{8}{5}, \frac{3 \pi}{2}\right) ,\left(\frac{8}{3}, \frac{3 \pi}{2}\right) C) directrix perpendicular to polar axis 8 left of pole center \left(\frac{8}{15}, 0\right) vertices \left(\frac{8}{5}, \pi\right) ,\left(\frac{8}{3}, 0\right) D) directrix parallel to polar axis 8 below pole center \left(\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(\frac{8}{3}, \frac{\pi}{2}\right) ,\left(\frac{8}{5}, \frac{3 \pi}{2}\right)

B) directrix parallel to polar axis 8 above pole center (−815,π2) \left(-\frac{8}{15}, \frac{\pi}{2}\right) vertices (−85,3π2) ,(83,3π2) \left(-\frac{8}{5}, \frac{3 \pi}{2}\right) ,\left(\frac{8}{3}, \frac{3 \pi}{2}\right)

Discuss the equation and graph it. - r = \frac { 4 } { 2 - \sin \theta } A) directrix perpendicular to polar axis 8 right of pole center \left(-\frac{8}{15}, 0\right) vertices \left(\frac{8}{3}, \pi\right) ,\left(\frac{8}{5}, 0\right) B) directrix parallel to polar axis 8 above pole center \left(-\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(-\frac{8}{5}, \frac{3 \pi}{2}\right) ,\left(\frac{8}{3}, \frac{3 \pi}{2}\right) C) directrix perpendicular to polar axis 8 left of pole center \left(\frac{8}{15}, 0\right) vertices \left(\frac{8}{5}, \pi\right) ,\left(\frac{8}{3}, 0\right) D) directrix parallel to polar axis 8 below pole center \left(\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(\frac{8}{3}, \frac{\pi}{2}\right) ,\left(\frac{8}{5}, \frac{3 \pi}{2}\right)

C) directrix perpendicular to polar axis 8 left of pole
center (815,0) \left(\frac{8}{15}, 0\right)
vertices (85,π) ,(83,0) \left(\frac{8}{5}, \pi\right) ,\left(\frac{8}{3}, 0\right)
Discuss the equation and graph it. - r = \frac { 4 } { 2 - \sin \theta } A) directrix perpendicular to polar axis 8 right of pole center \left(-\frac{8}{15}, 0\right) vertices \left(\frac{8}{3}, \pi\right) ,\left(\frac{8}{5}, 0\right) B) directrix parallel to polar axis 8 above pole center \left(-\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(-\frac{8}{5}, \frac{3 \pi}{2}\right) ,\left(\frac{8}{3}, \frac{3 \pi}{2}\right) C) directrix perpendicular to polar axis 8 left of pole center \left(\frac{8}{15}, 0\right) vertices \left(\frac{8}{5}, \pi\right) ,\left(\frac{8}{3}, 0\right) D) directrix parallel to polar axis 8 below pole center \left(\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(\frac{8}{3}, \frac{\pi}{2}\right) ,\left(\frac{8}{5}, \frac{3 \pi}{2}\right)

D) directrix parallel to polar axis 8 below pole center (815,π2) \left(\frac{8}{15}, \frac{\pi}{2}\right) vertices (83,π2) ,(85,3π2) \left(\frac{8}{3}, \frac{\pi}{2}\right) ,\left(\frac{8}{5}, \frac{3 \pi}{2}\right)
Discuss the equation and graph it. - r = \frac { 4 } { 2 - \sin \theta } A) directrix perpendicular to polar axis 8 right of pole center \left(-\frac{8}{15}, 0\right) vertices \left(\frac{8}{3}, \pi\right) ,\left(\frac{8}{5}, 0\right) B) directrix parallel to polar axis 8 above pole center \left(-\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(-\frac{8}{5}, \frac{3 \pi}{2}\right) ,\left(\frac{8}{3}, \frac{3 \pi}{2}\right) C) directrix perpendicular to polar axis 8 left of pole center \left(\frac{8}{15}, 0\right) vertices \left(\frac{8}{5}, \pi\right) ,\left(\frac{8}{3}, 0\right) D) directrix parallel to polar axis 8 below pole center \left(\frac{8}{15}, \frac{\pi}{2}\right) vertices \left(\frac{8}{3}, \frac{\pi}{2}\right) ,\left(\frac{8}{5}, \frac{3 \pi}{2}\right)

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Rotate the axes so that the new equation contains no xy-term. Discuss the new equati - 17x2−12xy+8y2−68x+24y−12=017 x ^ { 2 } - 12 x y + 8 y ^ { 2 } - 68 x + 24 y - 12 = 0


A) θ=26.6∘\theta = 26.6 ^ { \circ }
x′24+y′216=1\frac { x ^ { \prime 2 } } { 4 } + \frac { y ^ { \prime 2 } } { 16 } = 1
ellipse
center at (0,0) ( 0,0 )
major axis is y′y ^ { \prime } -axis
vertices at (0,±4) ( 0 , \pm 4 )

B) θ=63.4∘\theta = 63.4 ^ { \circ }
(x′−255) 216+(y′+455) 24=1\frac { \left( x ^ { \prime } - \frac { 2 \sqrt { 5 } } { 5 } \right) ^ { 2 } } { 16 } + \frac { \left( y ^ { \prime } + \frac { 4 \sqrt { 5 } } { 5 } \right) ^ { 2 } } { 4 } = 1
ellipse
center at (255,−455) \left( \frac { 2 \sqrt { 5 } } { 5 } , - \frac { 4 \sqrt { 5 } } { 5 } \right)
major axis is x′x ^ { \prime } -axis
vertices at (4+255,−455) \left( 4 + \frac { 2 \sqrt { 5 } } { 5 } , - \frac { 4 \sqrt { 5 } } { 5 } \right) and (−4+255,−455) \left( - 4 + \frac { 2 \sqrt { 5 } } { 5 } , - \frac { 4 \sqrt { 5 } } { 5 } \right)

C) θ=63.4∘\theta = 63.4 ^ { \circ }
x′2=−16y′x ^ { \prime 2 } = - 16 y ^ { \prime }
parabola
vertex at (0,0) ( 0,0 )
focus at (0,−4) ( 0 , - 4 )

D) θ=63.4∘\theta = 63.4 ^ { \circ }
x′216−y′24=1\frac { x ^ { \prime 2 } } { 16 } - \frac { y ^ { \prime 2 } } { 4 } = 1
hyperbola
center at (0,0) ( 0,0 )
transverse axis is the x′x ^ { \prime } -axis
vertices at (±4,0( \pm 4,0 )

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Write an equation for the hyperbola. - Write an equation for the hyperbola. - A) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1 B) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 9 } = 1 C) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1 D) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 16 } = 1


A) x24−y216=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1

B) y24−x29=1\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 9 } = 1

C) x24−y29=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1

D) y24−x216=1\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 16 } = 1

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Match the graph to its equation. - Match the graph to its equation. - A) \frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 9 } = 1 B) \frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 16 } = 1 C) \frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1 D) \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1


A) y216+x29=1\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 9 } = 1

B) x29−y216=1\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 16 } = 1

C) y216−x29=1\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1

D) x216+y29=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1

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Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola. - 25y2−16x2=40025 y ^ { 2 } - 16 x ^ { 2 } = 400


A) center at (0,0) ( 0,0 )
transverse axis is xx -axis
vertices: (−5,0) ,(5,0) ( - 5,0 ) , ( 5,0 )
foci: (−41,0) ,(41,0) ( - \sqrt { 41 } , 0 ) , ( \sqrt { 41 } , 0 )
asymptotes of y=−45y = - \frac { 4 } { 5 } and y=45y = \frac { 4 } { 5 }

B) center at (0,0) ( 0,0 )
transverse axis is xx -axis
vertices: (−4,0) ,(4,0) ( - 4,0 ) , ( 4,0 )
foci: (−5,0) ,(5,0) ( - 5,0 ) , ( 5,0 )
asymptotes of y=−45y = - \frac { 4 } { 5 } and y=45y = \frac { 4 } { 5 }

C) center at (0,0) ( 0,0 )
transverse axis is yy -axis
vertices: (0,−4) ,(0,4) ( 0 , - 4 ) , ( 0,4 )
foci: (−41,0) ,(41,0) ( - \sqrt { 41 } , 0 ) , ( \sqrt { 41 } , 0 )
asymptotes of y=−45y = - \frac { 4 } { 5 } and y=45y = \frac { 4 } { 5 }

D) center at (0,0) ( 0,0 )
transverse axis is y-axis
vertices at (0,−4) ( 0 , - 4 ) and (0,4) ( 0,4 )
foci at (0,−41) ( 0 , - \sqrt { 41 } ) and (0,41) ( 0 , \sqrt { 41 } )
asymptotes of y=−45\mathrm { y } = - \frac { 4 } { 5 } and y=45\mathrm { y } = \frac { 4 } { 5 }

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Find an equation for the parabola described.
-Vertex at (6, 1) ; focus at (6, 3) A) (y−1) 2=−12(x−6) ( y - 1 ) ^ { 2 } = - 12 ( x - 6 )
B) (y−1) 2=12(x−6) ( y - 1 ) ^ { 2 } = 12 ( x - 6 )
C) (x−6) 2=8(y−1) ( x - 6 ) ^ { 2 } = 8 ( y - 1 )
D) (x−6) 2=−8(y−1) ( x - 6 ) ^ { 2 } = - 8 ( y - 1 )

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Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17x2−12xy+8y2−68x+24y−12=017 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0 Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0 A) B)


A)
Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0 A) B)

B)
Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0 A) B)

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Find the vertex, focus, and directrix of the parabola. Graph the equation.
- y2+12y=4x−16y^{2}+12 y=4 x-16
Find the vertex, focus, and directrix of the parabola. Graph the equation. - y^{2}+12 y=4 x-16 A) vertex: ( - 5 , - 6 ) focus: ( - 4 , - 6 ) directrix: x = - 6 B) vertex: ( - 5 , - 6 ) focus: ( - 6 , - 6 ) directrix: x = - 4 C) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 7 ) directrix: y = - 5 D) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 5 ) directrix: y = - 7
A) vertex: (−5,−6) ( - 5 , - 6 )
focus: (−4,−6) ( - 4 , - 6 )
directrix: x=−6x = - 6
Find the vertex, focus, and directrix of the parabola. Graph the equation. - y^{2}+12 y=4 x-16 A) vertex: ( - 5 , - 6 ) focus: ( - 4 , - 6 ) directrix: x = - 6 B) vertex: ( - 5 , - 6 ) focus: ( - 6 , - 6 ) directrix: x = - 4 C) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 7 ) directrix: y = - 5 D) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 5 ) directrix: y = - 7

B) vertex: (−5,−6) ( - 5 , - 6 )
focus: (−6,−6) ( - 6 , - 6 )
directrix: x=−4x = - 4
Find the vertex, focus, and directrix of the parabola. Graph the equation. - y^{2}+12 y=4 x-16 A) vertex: ( - 5 , - 6 ) focus: ( - 4 , - 6 ) directrix: x = - 6 B) vertex: ( - 5 , - 6 ) focus: ( - 6 , - 6 ) directrix: x = - 4 C) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 7 ) directrix: y = - 5 D) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 5 ) directrix: y = - 7


C) vertex: (−5,−6) ( - 5 , - 6 )
focus: (−5,−7) ( - 5 , - 7 )
directrix: y=−5y = - 5
Find the vertex, focus, and directrix of the parabola. Graph the equation. - y^{2}+12 y=4 x-16 A) vertex: ( - 5 , - 6 ) focus: ( - 4 , - 6 ) directrix: x = - 6 B) vertex: ( - 5 , - 6 ) focus: ( - 6 , - 6 ) directrix: x = - 4 C) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 7 ) directrix: y = - 5 D) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 5 ) directrix: y = - 7

D) vertex: (−5,−6) ( - 5 , - 6 )
focus: (−5,−5) ( - 5 , - 5 )
directrix: y=−7y = - 7
Find the vertex, focus, and directrix of the parabola. Graph the equation. - y^{2}+12 y=4 x-16 A) vertex: ( - 5 , - 6 ) focus: ( - 4 , - 6 ) directrix: x = - 6 B) vertex: ( - 5 , - 6 ) focus: ( - 6 , - 6 ) directrix: x = - 4 C) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 7 ) directrix: y = - 5 D) vertex: ( - 5 , - 6 ) focus: ( - 5 , - 5 ) directrix: y = - 7


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Name the conic. -Name the conic. - A) circle B) parabola C) hyperbola D) ellipse


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

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Identify the equation without applying a rotation of axes. - 2x2+8xy+16y2−3x−2y+10=02 x ^ { 2 } + 8 x y + 16 y ^ { 2 } - 3 x - 2 y + 10 = 0


A) parabola
B) hyperbola
C) ellipse
D) not a conic

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Graph the curve whose parametric equations are given.
- x=3cos⁡t,y=2sin⁡t;0≤t≤2πx=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi

Graph the curve whose parametric equations are given. - x=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi A) B) C) D)
A)
Graph the curve whose parametric equations are given. - x=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi A) B) C) D)

B)
Graph the curve whose parametric equations are given. - x=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi A) B) C) D)

C)
Graph the curve whose parametric equations are given. - x=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi A) B) C) D)

D)
Graph the curve whose parametric equations are given. - x=3 \cos t, y=2 \sin t ; 0 \leq t \leq 2 \pi A) B) C) D)

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Find an equation for the hyperbola described. Graph the equation. -Center at (0,0) ;( 0,0 ) ; focus at (13,0) ( \sqrt { 13 } , 0 ) ;vertex at (2,0) ( 2,0 ) Find an equation for the hyperbola described. Graph the equation. -Center at ( 0,0 ) ; focus at ( \sqrt { 13 } , 0 ) ;vertex at ( 2,0 ) A) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1 B) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1 C) \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 D) \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1


A) y24−x225=1\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1
Find an equation for the hyperbola described. Graph the equation. -Center at ( 0,0 ) ; focus at ( \sqrt { 13 } , 0 ) ;vertex at ( 2,0 ) A) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1 B) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1 C) \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 D) \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1

B) x24−y225=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1
Find an equation for the hyperbola described. Graph the equation. -Center at ( 0,0 ) ; focus at ( \sqrt { 13 } , 0 ) ;vertex at ( 2,0 ) A) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1 B) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1 C) \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 D) \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1

C) y225−x24=1\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1
Find an equation for the hyperbola described. Graph the equation. -Center at ( 0,0 ) ; focus at ( \sqrt { 13 } , 0 ) ;vertex at ( 2,0 ) A) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1 B) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1 C) \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 D) \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1

D) x225−y24=1\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1
Find an equation for the hyperbola described. Graph the equation. -Center at ( 0,0 ) ; focus at ( \sqrt { 13 } , 0 ) ;vertex at ( 2,0 ) A) \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1 B) \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 25 } = 1 C) \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 D) \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1

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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 12.25 inches tall and stand 70 inches apart. Find the vertical distance from the Roadway to the cable at a point on the road 14 inches from the lowest point of the cable.


A) 2.16 in.
B) 1.76 in.
C) 7.84 in.
D) 1.96 in.

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Identify the equation without completing the square. - 2x2−4x+y+2=02 x ^ { 2 } - 4 x + y + 2 = 0


A) ellipse
B) hyperbola
C) parabola
D) not a conic

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Find an equation for the ellipse described. -Center at (0,0) ;( 0,0 ) ; focus at (0,5) ( 0,5 ) ; vertex at (0,−7) ( 0 , - 7 )


A) x224+y249=1\frac { x ^ { 2 } } { 24 } + \frac { y ^ { 2 } } { 49 } = 1

B) x225+y224=1\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 24 } = 1

C) x225+y249=1\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1

D) x249+y224=1\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 24 } = 1

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Determine the appropriate rotation formulas to use so that the new equation contains no xy-term. - 3x2+5xy+3y2−8x+8y=03 x ^ { 2 } + 5 x y + 3 y ^ { 2 } - 8 x + 8 y = 0


A) x=12x′−32y′x = \frac { 1 } { 2 } x ^ { \prime } - \frac { \sqrt { 3 } } { 2 } y ^ { \prime } and y=32x′+12y′y = \frac { \sqrt { 3 } } { 2 } x ^ { \prime } + \frac { 1 } { 2 } y ^ { \prime }
B) x=2+22x′−2−22y′x = \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } x ^ { \prime } - \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } y ^ { \prime } and y=2−22x′+2+22y′y = \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } x ^ { \prime } + \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } y ^ { \prime }
C) x=22(x′−y′) x = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } - y ^ { \prime } \right) and y=22(x′+y′) y = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } + y ^ { \prime } \right)
D) x=−y′x = - y ^ { \prime } and y=x′y = x ^ { \prime }

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Discuss the equation and graph it. - r=32+4sin⁡θr = \frac { 3 } { 2 + 4 \sin \theta } Discuss the equation and graph it. - r = \frac { 3 } { 2 + 4 \sin \theta } A) ellipse, directrix parallel to the polar axis \frac{3}{2} unit below the pole vertices \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right) B) ellipse, directrix perpendicular to the polar axis \frac{3}{2} unit left of the pole vertices \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right) C) hyperbola; directrix parallel to the polar axis \frac{3}{4} unit above the pole vertices \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) D) hyperbola, directrix perpendicular to the polar axis \frac{3}{4} unit right of the pole vertices \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)


A) ellipse, directrix parallel to the polar axis 32 \frac{3}{2} unit below the pole vertices (32,π2) ,(12,3π2) \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right)
Discuss the equation and graph it. - r = \frac { 3 } { 2 + 4 \sin \theta } A) ellipse, directrix parallel to the polar axis \frac{3}{2} unit below the pole vertices \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right) B) ellipse, directrix perpendicular to the polar axis \frac{3}{2} unit left of the pole vertices \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right) C) hyperbola; directrix parallel to the polar axis \frac{3}{4} unit above the pole vertices \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) D) hyperbola, directrix perpendicular to the polar axis \frac{3}{4} unit right of the pole vertices \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)

B) ellipse, directrix perpendicular to the polar axis 32 \frac{3}{2} unit left of the pole vertices (32,0) ,(12,π) \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right)
Discuss the equation and graph it. - r = \frac { 3 } { 2 + 4 \sin \theta } A) ellipse, directrix parallel to the polar axis \frac{3}{2} unit below the pole vertices \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right) B) ellipse, directrix perpendicular to the polar axis \frac{3}{2} unit left of the pole vertices \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right) C) hyperbola; directrix parallel to the polar axis \frac{3}{4} unit above the pole vertices \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) D) hyperbola, directrix perpendicular to the polar axis \frac{3}{4} unit right of the pole vertices \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)

C) hyperbola; directrix parallel to the polar axis 34 \frac{3}{4} unit above the pole vertices (12,π2) ,(−32,3π2) \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right)
Discuss the equation and graph it. - r = \frac { 3 } { 2 + 4 \sin \theta } A) ellipse, directrix parallel to the polar axis \frac{3}{2} unit below the pole vertices \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right) B) ellipse, directrix perpendicular to the polar axis \frac{3}{2} unit left of the pole vertices \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right) C) hyperbola; directrix parallel to the polar axis \frac{3}{4} unit above the pole vertices \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) D) hyperbola, directrix perpendicular to the polar axis \frac{3}{4} unit right of the pole vertices \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)

D) hyperbola, directrix perpendicular to the polar axis 34 \frac{3}{4} unit right of the pole vertices (12,0) ,(−32,π) \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)
Discuss the equation and graph it. - r = \frac { 3 } { 2 + 4 \sin \theta } A) ellipse, directrix parallel to the polar axis \frac{3}{2} unit below the pole vertices \left(\frac{3}{2}, \frac{\pi}{2}\right) ,\left(\frac{1}{2}, \frac{3 \pi}{2}\right) B) ellipse, directrix perpendicular to the polar axis \frac{3}{2} unit left of the pole vertices \left(\frac{3}{2}, 0\right) ,\left(\frac{1}{2}, \pi\right) C) hyperbola; directrix parallel to the polar axis \frac{3}{4} unit above the pole vertices \left(\frac{1}{2}, \frac{\pi}{2}\right) ,\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) D) hyperbola, directrix perpendicular to the polar axis \frac{3}{4} unit right of the pole vertices \left(\frac{1}{2}, 0\right) ,\left(-\frac{3}{2}, \pi\right)

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Match the equation to the graph. - (y−2) 2=6(x−2) ( y - 2 ) ^ { 2 } = 6 ( x - 2 )


A)
Match the equation to the graph. - ( y - 2 ) ^ { 2 } = 6 ( x - 2 ) A) B) C) D)

B)
Match the equation to the graph. - ( y - 2 ) ^ { 2 } = 6 ( x - 2 ) A) B) C) D)
C)
Match the equation to the graph. - ( y - 2 ) ^ { 2 } = 6 ( x - 2 ) A) B) C) D)

D)
Match the equation to the graph. - ( y - 2 ) ^ { 2 } = 6 ( x - 2 ) A) B) C) D)

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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 stand 50 inches apart. At a point between the towers and 15 inches along the road from the Base of one tower, the cable is 1 inches above the roadway. Find the height of the towers.


A) 6.75 in.
B) 5.75 in.
C) 6.25 in.
D) 8.25 in.

Correct Answer

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