Exam 11: Analytic Geometry

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

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Graph the conic section. - y212y+4x+48=0y ^ { 2 } - 12 y + 4 x + 48 = 0  Graph the conic section. - y ^ { 2 } - 12 y + 4 x + 48 = 0

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Determine the two equations necessary to graph the ellipse with a graphing calculator. - x264+(y2)281=1\frac { x ^ { 2 } } { 64 } + \frac { ( y - 2 ) ^ { 2 } } { 81 } = 1

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Provide an appropriate response. -The eccentricity of ellipse A\mathrm { A } is eA=0.18\mathrm { e } _ { \mathrm { A } } = 0.18 and the eccentricity of ellipse B\mathrm { B } is eB=0.82\mathrm { eB } = 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)2y + 2 = 2 ( x + 15 ) ^ { 2 }

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

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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 \times 10 ^ { 6 } m.

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Write an equation for the hyperbola. -The roof of a building is in the shape of the hyperbola y22x2=56y ^ { 2 } - 2 x ^ { 2 } = 56 , where xx and yy are in meters. Refer to the figure and determine the height, hh , of the outside walls. A=3 mA = 3 \mathrm {~m}  Write an equation for the hyperbola. -The roof of a building is in the shape of the hyperbola  y ^ { 2 } - 2 x ^ { 2 } = 56 , where  x  and  y  are in meters. Refer to the figure and determine the height,  h , of the outside walls.  A = 3 \mathrm {~m}

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

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Graph the parabola. - x=(y3)2x=(y-3)^{2}  Graph the parabola. - x=(y-3)^{2}

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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 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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Graph the conic section. - 9x216y236x96y252=09 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

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Find the center, foci, and asymptotes of the hyperbola. - x2y2=32x ^ { 2 } - y ^ { 2 } = 32

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Match the equation of the parabola with the appropriate description. - y=6x26x+4y = 6 x ^ { 2 } - 6 x + 4

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

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

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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 } at the center. The distance between the two foci is 92ft92 \mathrm { ft } . What is the equation for the hyperbola?

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

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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

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

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