Exam 9: Analytic Geometry

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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.

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Graph the hyperbola. - x24−y236=1\frac{x^{2}}{4}-\frac{y^{2}}{36}=1  Graph the hyperbola. - \frac{x^{2}}{4}-\frac{y^{2}}{36}=1

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Determine the appropriate rotation formulas to use so that the new equation contains no xy-term. - 11x2−4xy+7y2−8x+8y=011 x ^ { 2 } - 4 x y + 7 y ^ { 2 } - 8 x + 8 y = 0

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Find parametric equations for the rectangular equation. - y=3x2+6y = 3 x ^ { 2 } + 6

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Match the equation to its graph. - x2=−12yx ^ { 2 } = - 12 y

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Find a polar equation for the conic. A focus is at the pole. - e=5\mathrm { e } = 5 ; directrix is perpendicular to the polar axis 3 to the right of the pole

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

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Find a rectangular equation for the plane curve defined by the parametric equations. - x=5cos⁡t,y=−2sin⁡t;0≤t≤2πx = 5 \cos t , y = - 2 \sin t ; 0 \leq t \leq 2 \pi

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Find the center, foci, and vertices of the ellipse. - x216+y264=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1

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

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

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Find an equation for the hyperbola described. -center at (9,4)( 9,4 ) ; focus at (7,4)( 7,4 ) ; vertex at (8,4)( 8,4 )

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Write an equation for the graph. -Write an equation for the graph. -

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Discuss the equation and graph it. - r=63−3cos⁡θr=\frac{6}{3-3 \cos \theta}  Discuss the equation and graph it. - r=\frac{6}{3-3 \cos \theta}

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Find the vertex, focus, and directrix of the parabola with the given equation. - (x+1)2=12(y−4)( x + 1 ) ^ { 2 } = 12 ( y - 4 )

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Match the equation to its graph. - y2=13xy ^ { 2 } = 13 x

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Graph the hyperbola. - 25y2−4x2=10025 y^{2}-4 x^{2}=100  Graph the hyperbola. - 25 y^{2}-4 x^{2}=100

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Solve the problem. -Find parametric equations for an object that moves along the ellipse 1 with the motion described. x29+y281=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 81 } = 1 The motion begins at (3,0)( 3,0 ) , is counterclockwise, and requires 5 seconds for a complete revolution.

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Solve the problem. -The roof of a building is in the shape of the hyperbola y2−x2=49y ^ { 2 } - x ^ { 2 } = 49 , where xx and yy are in meters. Determine the distance, ww , the outside walls are apart, if the height of each wall is 12 m12 \mathrm {~m} .  Solve the problem. -The roof of a building is in the shape of the hyperbola  y ^ { 2 } - x ^ { 2 } = 49 , where  x  and  y  are in meters. Determine the distance,  w , the outside walls are apart, if the height of each wall is  12 \mathrm {~m} .

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Rotate the axes so that the new equation contains no xy-term. Discuss the new equation. - x2+xy+y2−3y−6=0x ^ { 2 } + x y + y ^ { 2 } - 3 y - 6 = 0

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