Practice Test B

  1. Find the standard form of the equation of the parabola with focus (10, 0) and directrix x=10.

    1. y2=10x

    2. y2=40x

    3. x2=40y

    4. y2=40x

  2. Convert the equation y24y5x+24=0 to the standard form for a parabola by completing the square.

    1. (y+2)2=5(x4)

    2. (y2)2=5(x4)

    3. (y+2)2=5(x4)

    4. (y2)2=5(x+4)

  3. Find the focus and directrix of the parabola with equation x=7y2.

    1. focus: (128, 0); directrix: x=128

    2. focus: (0,128); directrix: y=128

    3. focus: (128, 0); directrix: x=128

    4. focus: (17, 0) directrix: x=17

  4. Graph the parabola with equation y2=7x.

  5. Find the vertex, focus, and directrix of the parabola with equation (x3)2=12(y1).

    1. vertex: (3, 1); focus: (3, 4); directrix: y=2

    2. vertex: (3, 1); focus: (3, 2); directrix: y=4

    3. vertex: (3, 1); focus: (3, 2); directrix: x=4

    4. vertex: (1, 3); focus: (1, 6); directrix: y=0

  6. The parabolic arch of a bridge has a 180-foot base and a height of 25 feet. Find the height of the arch 45 feet from the center of the base.

    1. 12.5 ft

    2. 6.25 ft

    3. 16.7 ft

    4. 18.75 ft

  7. Find an equation of the parabola with vertex (2, 1) and directrix x=2.

    1. (y1)2=16(x+2)

    2. (y1)2=16(x+2)

    3. (y+2)2=16(x1)

    4. (y+2)2=16(x1)

In Problems 8–10, find the standard form of the equation of the ellipse satisfying the given conditions.

  1. Foci: (3, 0), (3, 0); vertices: (5, 0), (5, 0)

    1. x29+y216=1

    2. x225+y216=1

    3. x29+y225=1

    4. x216+y225=1

  2. Foci: (0, 3), (0, 3); y-intercepts: 7, 7

    1. x249+y240=1

    2. x29+y249=1

    3. x29+y240=1

    4. x240+y249=1

  3. Major axis vertical with length 16; length of minor axis=8; center: (0, 0)

    1. x216+y264=1

    2. x264+y2256=1

    3. x28+y264=1

    4. x264+y216=1

  4. Graph the ellipse with equation 9(x1)2+4(y2)2=36.

  5. Find the vertices and foci of the hyperbola with equation x2121y24=1.

    1. vertices: (11, 0), (11, 0); foci: (2, 0), (2, 0)

    2. vertices: (0, 1), (0, 11); foci: (55, 0)(55, 0)

    3. vertices: (11, 0), (11, 0); foci: (55, 0)(55, 0)

    4. vertices: (2, 0), (2, 0); foci: (55, 0)(55, 0)

  6. Graph the hyperbola with equation x225y24=1.

  7. Find the standard form of the equation of the hyperbola with foci (0, 10), (0, 10) and vertices (0, 5), (0, 5).

    1. y225x2100=1

    2. y225x275=1

    3. x225y2100=1

    4. x225y275=1

In Problems 15 and 16, convert the equation to the standard form for a hyperbola by completing the square on x and y.

  1. y24x22y16x19=0

    1. (y2)24(x+4)2=1

    2. (x1)24(y+2)2=1

    3. (y1)24(x+2)2=1

    4. (x+2)2(y1)24=1

  2. 4y29x216y36x56=0

    1. (y2)24(x+2)29=1

    2. (x2)24(y+2)29=1

    3. (y2)29(x+2)24=1

    4. (y+2)29(x2)24=1

In Problems 17–20, identify the conic section.

  1. 3x2+2y26x+4y1=0

    1. parabola

    2. circle

    3. ellipse

    4. hyperbola

  2. x2+2x+4y+5=0

    1. parabola

    2. circle

    3. ellipse

    4. hyperbola

  3. 4x2y2+16x+2y+11=0

    1. parabola

    2. circle

    3. ellipse

    4. hyperbola

  4. 5y23x2+20y6x+2=0

    1. parabola

    2. circle

    3. ellipse

    4. hyperbola

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