Exam 10: Analytic Geometry

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Graph the parabola. - x=−2(y+2)2+1x = - 2 ( y + 2 ) ^ { 2 } + 1  Graph the parabola. - x = - 2 ( y + 2 ) ^ { 2 } + 1

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Choose the equation that matches the graph. -Choose the equation that matches the graph. -

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Identify the type of graph. - 2x2−4x+5y2+30y−43=02 x ^ { 2 } - 4 x + 5 y ^ { 2 } + 30 y - 43 = 0

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Solve the problem. -A cross-section of an irrigation canal is a parabola. If the surface of the water is 40ft40 \mathrm { ft } wide and the canal is 26ft26 \mathrm { ft } deep at the center, how deep is it 1ft1 \mathrm { ft } from the edge?

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Write an equation for the ellipse. -minor axis from (−7,−1)( - 7 , - 1 ) to (−1,−1)( - 1 , - 1 ) ; major axis from (−4,−7)( - 4 , - 7 ) to (−4,5)( - 4,5 )

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Solve the problem. -A rectangular board is 8 by 9 units. The foci of an ellipse are located to produce the largest area. A string is connected to the foci and pulled taut by a pencil in order to draw the ellipse. Find the Length of the string.

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Match the equation of the parabola with the appropriate description. - y+10=2(x−15)2y + 10 = 2 ( x - 15 ) ^ { 2 }

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Graph the hyperbola. - 4y2−36x2=1444 y^{2}-36 x^{2}=144  Graph the hyperbola. - 4 y^{2}-36 x^{2}=144

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Write the word or phrase that best completes each statement or answers the question. -  Discuss the graph of the equation x2a2+y2b2=1 if a=b. \text { Discuss the graph of the equation } \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 \text { if } a = b \text {. }

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Give the focus, directrix, and axis for the parabola. - −124x2=y- \frac { 1 } { 24 } x ^ { 2 } = y

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Solve the problem. -A building has an entry the shape of a parabolic arch 4 ft high and 28 ft wide at the base. Find an equation for the parabola if the vertex is put at the origin of the coordinate system. Solve the problem. -A building has an entry the shape of a parabolic arch 4 ft high and 28 ft wide at the base. Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

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When a satellite is near Earth, its orbital trajectory may trace out a hyperbola, a parabola, or an ellipse. The type of trajectory depends on the satellite's velocity VV in meters per second. It will be hyperbolic if V>kDV > \frac { k } { \sqrt { D } } , parabolic if V=V = kD\frac { k } { \sqrt { D } } , and elliptical if V<kDV < \frac { k } { \sqrt { D } } , where k=2.82×107k = 2.82 \times 10 ^ { 7 } is a constant and D\mathrm { D } is the distance in meters from the satellite to the center of Earth. Solve the problem. -If a satellite is scheduled to leave Earthʹs 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 4881 meters per second to escape Earthʹs Gravitational influence when D = 194 × 106 m.

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Find the eccentricity of the conic section shown in the graph. -Find the eccentricity of the conic section shown in the graph. -

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Write the word or phrase that best completes each statement or answers the question. -Explain why the graph of an ellipse does not satisfy the conditions for the graph of a function.

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Give the focus, directrix, and axis for the parabola. - (x+2)2=−28(y−3)( x + 2 ) ^ { 2 } = - 28 ( y - 3 )

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Write an equation for the parabola. -vertex (7,−9)( 7 , - 9 ) , focus (10,−9)( 10 , - 9 )

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Graph the conic section. -Graph the conic section. -

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Graph the hyperbola. - y225−x236=1\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 36 } = 1  Graph the hyperbola. - \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 36 } = 1

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Find the eccentricity of the hyperbola. - x2−16y2=1x ^ { 2 } - 16 y ^ { 2 } = 1

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Fill in the blanks to complete the statement. The graph of P(x)=2(x−8)2−4P ( x ) = 2 ( x - 8 ) ^ { 2 } - 4 is a parabola with axis of symmetryــــــــــــ and vertexــــــــــــــــــــ

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