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Find the vertices and locate the foci for the hyperbola whose equation is given

Question : Find the vertices and locate the foci for the hyperbola whose equation is given : 2151874

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Find the vertices and locate the foci for the hyperbola whose equation is given.

1) (x2/121) - (y2/36) = 1

A) vertices: (-6, 0), (6, 0)

foci: (- √(157), 0), (√(157), 0)

B) vertices: (0, -11), (0, 11)

foci: (- √(157), 0), (√(157), 0)

C) vertices: (-11, 0), (11, 0)

foci: (- √(157), 0), (√(157), 0)

D) vertices: (-11, 0), (11, 0)

foci: (-6, 0), (6, 0)

2) (y2/49) - (x2/81) = 1

A) vertices: (-9, 0), (9, 0)

foci: (- √(130), 0), (√(130), 0)

B) vertices: (-7, 0), (7, 0)

foci: (-9, 0), (9, 0)

C) vertices: (0, -7), (0, 7)

foci: (- √(130), 0), (√(130), 0)

D) vertices: (0, -7), (0, 7)

foci: (0, - √(130)), (0, √(130))

3) 25x2 - 16y2 = 400

A) vertices: (0, -4), (0, 4)

foci: (0, -√(41)), (0, √(41))

B) vertices: (-4, 0), (4, 0)

foci: (-√(41), 0), (√(41), 0)

C) vertices: (-5, 0), (5, 0)

foci: (-√(41), 0), (√(41), 0)

D) vertices: (-4, 0), (4, 0)

foci: (-3, 0), (3, 0)

4) 9y2 - 100x2 = 900

A) vertices: (-3, 0), (3, 0)

foci: (-√(91), 0), (√(91), 0)

B) vertices: (-10, 0), (10, 0)

foci: (-√(109), 0), (√(109), 0)

C) vertices: (0, -3), (0, 3)

foci: (0, -√(109)), (0, √(109))

D) vertices: (0, -10), (0, 10)

foci: (0, -√(109)), (0, √(109))

5) y = ± √(x2 - 9)

A) vertices: (-9, 0), (9, 0)

foci: (-3√(2), 0), (3√(2), 0)

B) vertices: (0, -3), (0, 3)

foci: (0, -3√(2)), (0, 3√(2))

C) vertices: (-9, 0), (9, 0)

foci: (-3, 0), (3, 0)

D) vertices: (-3, 0), (3, 0)

foci: (-3√(2), 0), (3√(2), 0)

Match the equation to the graph.

6) (x2/4) - (y2/9) = 1

A)

B)

C)

D)

7) (y2/4) - (x2/16) = 1

A)

B)

C)

D)

Find the standard form of the equation of the hyperbola satisfying the given conditions.

8) Foci: (-8, 0), (8, 0); vertices: (-7, 0), (7, 0)

A) (y2/49) - (x2/15) = 1

B) (y2/49) - (x2/64) = 1

C) (x2/49) - (y2/64) = 1

D) (x2/49) - (y2/15) = 1

9) Foci: (0, -9), (0, 9); vertices: (0, -4), (0, 4)

A) (x2/16) - (y2/81) = 1

B) (x2/16) - (y2/65) = 1

C) (y2/16) - (x2/81) = 1

D) (y2/16) - (x2/65) = 1

10) Endpoints of transverse axis: (0, -6), (0, 6); asymptote: y = (3/10)x

A) (y2/400) - (x2/36) = 1

B) (y2/100) - (x2/9) = 1

C) (y2/36) - (x2/100) = 1

D) (y2/36) - (x2/400) = 1

11) Endpoints of transverse axis: (-5, 0), (5, 0); foci: (-9, 0), (-9, 0)

A) (x2/25) - (y2/81) = 1

B) (x2/81) - (y2/25) = 1

C) (x2/56) - (y2/25) = 1

D) (x2/25) - (y2/56) = 1

12) Center: (5, 6); Focus: (-2, 6); Vertex: (4, 6)

A) (x - 5)2 - ((y - 6)2/48) = 1

B) ((x - 5)2/48) - (y - 6)2 = 1

C) ((x - 6)2/48) - (y - 5)2 = 1

D) (x - 6)2 - ((y - 5)2/48) = 1

Find the standard form of the equation of the hyperbola.

13)

A) (x2/16) - (y2/9) = 1

B) (y2/16) - (x2/9) = 1

C) (x2/9) - (y2/16) = 1

D) (y2/9) - (x2/16) = 1

14)

A) (x2/25) - (y2/4) = 1

B) (y2/4) - (x2/25) = 1

C) (x2/4) - (y2/25) = 1

D) (y2/25) - (x2/4) = 1

15)

A) ((y - 1)2/25) - ((x - 1)2/4) = 1

B) ((x - 1)2/4) - ((y - 1)2/25) = 1

C) ((x - 1)2/25) - ((y - 1)2/4) = 1

D) ((y - 1)2/4) - ((x - 1)2/25) = 1

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

16) x2 - y2 + 4x + 2y + 2 = 0

A) (y + 2)2 - (x - 1)2 = 1

B) (x + 2)2 - (y - 1)2 = 1

C) (x + 2)2 + (y - 1)2 = 1

D) ((y + 2)2/4) - ((x - 1)2/16) = 1

17) y2 - 25x2 - 4y + 50x - 46 = 0

A) ((y - 4)2/25) - (x - 2)2 = 1

B) (x - 1)2 - ((y - 2)2/25) = 1

C) ((x - 2)2/25) - (y - 1)2 = 1

D) ((y - 2)2/25) - (x - 1)2 = 1

18) 4x2 - 16y2 + 8x - 32y - 76 = 0

A) ((x + 1)2/4) - ((y + 1)2/16) = 1

B) ((x + 1)2/16) - ((y - 1)2/4) = 1

C) ((x - 1)2/16) - ((y + 1)2/4) = 1

D) ((x + 1)2/16) - ((y + 1)2/4) = 1

19) 4y2 - 9x2 - 16y + 36x - 56 = 0

A) ((x + 2)2/4) - ((y + 2)2/9) = 1

B) ((y + 2)2/9) - ((x + 2)2/4) = 1

C) ((y - 2)2/4) - ((x - 2)2/9) = 1

D) ((y - 2)2/9) - ((x - 2)2/4) = 1

Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes.

20) (x2/4) - (y2/9) = 1

A) Asymptotes: y = ± (2/3)x

B) Asymptotes: y = ± (3/2)x

C) Asymptotes: y = ± (2/3)x

D) Asymptotes: y = ± (3/2)x

21) (y2/4) - (x2/9) = 1

A) Asymptotes: y = ± (3/2)x

B) Asymptotes: y = ± (2/3)x

C) Asymptotes: y = ± (3/2)x

D) Asymptotes: y = ± (2/3)x

Solution
5 (1 Ratings )

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