Exam 8: Analytic Geometry in Two and Three Dimensions

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Solve the problem. -A rectangular board is 14 by 20 units. How far from the short side of the board will the foci be located to determine the largest elliptical tabletop?

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A quadratic equation has been transformed using the rotation equations x = 0.8u - 0.6v and y = 0.6u + 0.8v. Convert the given point from (u, v) coordinates back to (x, y) coordinates. -hyperbola foci (±3,0)( \pm 3,0 )

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Find an equation that matches the parabola's graph. -Find an equation that matches the parabola's graph. -

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Graph the ellipse. - x225+y29=1\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1  Graph the ellipse. - \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1

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Find the values of e, a, b, and c. - r=244+2sinθr = \frac { 24 } { 4 + 2 \sin \theta }

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Find the vertices and foci of the hyperbola. - x24y212=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 12 } = 1

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Match the polar equation with its graph. - r=84sinθr = \frac { 8 } { 4 - \sin \theta }

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Find the eccentricity of the ellipse. - x2+3y2=15x ^ { 2 } + 3 y ^ { 2 } = 15

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Solve the problem. -A railroad tunnel is shaped like a semiellipse. The height of the tunnel at the center is 28ft28 \mathrm { ft } and the vertical clearance must be 26ft26 \mathrm { ft } at a point 18ft18 \mathrm { ft } from the center. Find an equation for the ellipse.  Solve the problem. -A railroad tunnel is shaped like a semiellipse. The height of the tunnel at the center is  28 \mathrm { ft }  and the vertical clearance must be  26 \mathrm { ft }  at a point  18 \mathrm { ft }  from the center. Find an equation for the ellipse.

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Find the standard form of the equation of the parabola. -Focus at (0,3)( 0,3 ) , directrix y=3y = - 3

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Find the eccentricity of the hyperbola. - x22y2=6x ^ { 2 } - 2 y ^ { 2 } = 6

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Compute the distance between the points. - (4,4,6),(4,5,10)( 4,4 , - 6 ) , ( - 4,5,10 )

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Match the polar equation with its graph. - r=128+6cosθr = \frac { 12 } { 8 + 6 \cos \theta }

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Graph the ellipse. - 25x2+25y2=62525 x^{2}+25 y^{2}=625  Graph the ellipse. - 25 x^{2}+25 y^{2}=625

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Find parametric equations for the line described below. -The line through the point P(3,4,2)P ( - 3,4 , - 2 ) in the direction of the vector 8,4,5\langle - 8,4 , - 5 \rangle

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Match the given graph with its equation. -Match the given graph with its equation. -

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Graph the hyperbola. - 25x24y2=10025 x^{2}-4 y^{2}=100  Graph the hyperbola. - 25 x^{2}-4 y^{2}=100

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Compute the distance between the points. - (6,2,9),(x,y,z)( 6 , - 2,9 ) , ( x , y , z )

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Find the midpoint of the segment PQ. - P(9,10,3),Q(1,15,7)\mathrm { P } ( - 9,10 , - 3 ) , \mathrm { Q } ( 1,15,7 )

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Evaluate the expression. - r=7,5,3,v=2,4,5,w=1,7,5\mathbf { r } = \langle 7 , - 5 , - 3 \rangle , \mathbf { v } = \langle 2,4 , - 5 \rangle , \mathbf { w } = \langle 1,7,5 \rangle vw\mathbf { v } \cdot \mathbf { w }

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