Exam 8: Analytic Geometry in Two and Three Dimensions

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

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Find the values of e, a, b, and c. - r=215−2cos⁡θr = \frac { 21 } { 5 - 2 \cos \theta }

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Find an equation in standard form for the hyperbola that satisfies the given conditions. -Vertices at (0,±8)( 0 , \pm 8 ) , asymptotes at y=±25xy = \pm \frac { 2 } { 5 } x

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

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Find the standard form of the equation of the parabola. -Vertex at the origin, focus at (8,0)( 8,0 )

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Find the linear equations that can be used to convert an (x, y) equation to a (u, v) equation using the given angle of rotation - θ=tan⁡−1(43)\theta = \tan ^ { - 1 } \left( \frac { 4 } { 3 } \right)

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Convert the equation to Cartesian form. - r=−6618+12sin⁡θr = \frac { - 66 } { 18 + 12 \sin \theta }

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Find the eccentricity of the hyperbola. - x2−y2=49x ^ { 2 } - y ^ { 2 } = 49

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Match the polar equation with its graph. - r=104−6sin⁡θr = \frac { 10 } { 4 - 6 \sin \theta }

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

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Find an equation in standard form for the ellipse that satisfies the given conditions. -Minor axis endpoints (±4,0)( \pm 4,0 ) , major axis length 22

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Use the discriminant to decide whether the equation represents a parabola, an ellipse, or a hyperbola. - x2+4xy+6y2−2x+4y=0x ^ { 2 } + 4 x y + 6 y ^ { 2 } - 2 x + 4 y = 0

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Use the given substitutions to convert the equation to a (u, v) equation with no cross-product term. - 5x2−6xy+5y2−8=0;5 x ^ { 2 } - 6 x y + 5 y ^ { 2 } - 8 = 0 ; use x=u/2−v/2x = u / \sqrt { 2 } - v / \sqrt { 2 } and y=u/2+v/2y = u / \sqrt { 2 } + v / \sqrt { 2 }

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Find the vertices and foci of the conic section without axis rotation by analyzing the graph geometrically in the xy-plane. -The hyperbola xy=−9x y = - 9

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Find the midpoint of the segment PQ. - P(4,−4,−8),Q(10,−1,−6)\mathrm { P } ( 4 , - 4 , - 8 ) , \mathrm { Q } ( 10 , - 1 , - 6 )

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Match the polar equation with its graph. - r=168−7cos⁡θr = \frac { 16 } { 8 - 7 \cos \theta }

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Solve the problem. -An elliptical riding path is to be built on a rectangular piece of property that measures 8 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides Solve the problem. -An elliptical riding path is to be built on a rectangular piece of property that measures 8 mi by 4 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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Solve the problem. - Solve the problem. -  A tunnel is in the shape of a parabola. The maximum height is  30 \mathrm {~m}  and it is  18 \mathrm {~m}  wide at the base. What is the vertical clearance  2 \mathrm {~m}  from the edge of the tunnel? A tunnel is in the shape of a parabola. The maximum height is 30 m30 \mathrm {~m} and it is 18 m18 \mathrm {~m} wide at the base. What is the vertical clearance 2 m2 \mathrm {~m} from the edge of the tunnel?

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

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Graph the parabola. - (x+2)2=4(y−5)(x+2)^{2}=4(y-5)  Graph the parabola. - (x+2)^{2}=4(y-5)

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