Exam 10: Conic Sections and Analytic Geometry

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Tech: Parametric Equations -Witch of Agnesi: x=3cott,y=3sin2t,0t<2πx = 3 \cot t , y = 3 \sin ^ { 2 } t , 0 \leq t < 2 \pi  Tech: Parametric Equations -Witch of Agnesi:  x = 3 \cot t , y = 3 \sin ^ { 2 } t , 0 \leq t < 2 \pi

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Solve Apps: Conic Sections in Polar Coordinates -Halley's comet has an elliptical orbit with the sun at one focus. Its orbit shown below is given approximately by r=10.751+0.838sinθr = \frac { 10.75 } { 1 + 0.838 \sin \theta } . In the formula, rr is measured in astronomical units. (One astronomical unit is the average distance from Earth to the sun, approximately 93 million miles.) Find the distance from Halley's comet to the sun at its shortest distance from the sun. Round to the nearest hundredth of an astronomical unit and the nearest million miles.  Solve Apps: Conic Sections in Polar Coordinates -Halley's comet has an elliptical orbit with the sun at one focus. Its orbit shown below is given approximately by  r = \frac { 10.75 } { 1 + 0.838 \sin \theta } . In the formula,  r  is measured in astronomical units. (One astronomical unit is the average distance from Earth to the sun, approximately 93 million miles.) Find the distance from Halley's comet to the sun at its shortest distance from the sun. Round to the nearest hundredth of an astronomical unit and the nearest million miles.

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Additional Concepts - {x225+y29=1y=3\left\{ \begin{array} { l } \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1 \\y = 3\end{array} \right.  Additional Concepts - \left\{ \begin{array} { l }  \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1 \\ y = 3 \end{array} \right.

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Use the center, vertices, and asymptotes to graph the hyperbola. - (y2)29(x1)225=1\frac { ( y - 2 ) ^ { 2 } } { 9 } - \frac { ( x - 1 ) ^ { 2 } } { 25 } = 1  Use the center, vertices, and asymptotes to graph the hyperbola. - \frac { ( y - 2 ) ^ { 2 } } { 9 } - \frac { ( x - 1 ) ^ { 2 } } { 25 } = 1

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Solve Applied Problems Involving Parabolas -A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 184 feet and a maximum height of 40 feet. Find the height of the arch at 15 feet from its center.

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Graph Hyperbolas Not Centered at the Origin - (x3)24(y1)29=1\frac { ( x - 3 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1

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Write Equations of Rotated Conics in Standard Form - 4x24xy+y26x+12=04 x ^ { 2 } - 4 x y + y ^ { 2 } - 6 x + 12 = 0

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Identify Conics Without Rotating Axes - 14x2+123xy+2y229=014 x ^ { 2 } + 12 \sqrt { 3 } x y + 2 y ^ { 2 } - 29 = 0

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Graph the ellipse and locate the foci. - 9x2=14416y29 x^{2}=144-16 y^{2}  Graph the ellipse and locate the foci. - 9 x^{2}=144-16 y^{2}

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The Hyperbola 1 Locate a Hyperbola's Vertices and Foci - 4y216x2=644 y ^ { 2 } - 16 x ^ { 2 } = 64

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Graph the semi-ellipse. - y=2516x2y=-\sqrt{25-16 x^{2}}  Graph the semi-ellipse. - y=-\sqrt{25-16 x^{2}}

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Use point plotting to graph the plane curve described by the given parametric equations. - x=3t,y=t+3;2t3x=3 t, y=t+3 ;-2 \leq t \leq 3  Use point plotting to graph the plane curve described by the given parametric equations. - x=3 t, y=t+3 ;-2 \leq t \leq 3

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Rotation of Axes 1 Identify Conics Without Completing the Square - 3y24x4y=03 y ^ { 2 } - 4 x - 4 y = 0

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The Hyperbola 1 Locate a Hyperbola's Vertices and Foci - 81x2100y2=810081 x ^ { 2 } - 100 y ^ { 2 } = 8100

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Tech: Rotation of Axes - x224xy+8y2=36x^{2}-24 x y+8 y^{2}=36  Tech: Rotation of Axes - x^{2}-24 x y+8 y^{2}=36

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Additional Concepts - y=x22x+7y = x ^ { 2 } - 2 x + 7

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Use Rotation of Axes Formulas - xy+16=0;θ=45\mathrm { xy } + 16 = 0 ; \quad \theta = 45 ^ { \circ }

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Write the appropriate rotation formulas so that in a rotated system the equation has no x'y'-term. - x2+2xy+y28x+8y=0x ^ { 2 } + 2 x y + y ^ { 2 } - 8 x + 8 y = 0

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Use Rotation of Axes Formulas - x2+2xy+y28x+8y=0;θ=45x ^ { 2 } + 2 x y + y ^ { 2 } - 8 x + 8 y = 0 ; \quad \theta = 45 ^ { \circ }

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Write Equations of Hyperbolas in Standard Form -Center: (6,1)( 6,1 ) ; Focus: (1,1)( - 1,1 ) ; Vertex: (5,1)( 5,1 )

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