College Algebra Problems With Answers sample 10 : Equation of Hyperbola

College algebra problems on the equations of hyperbolas are presented. Detailed solutions are at the bottom of the page.

Problem 1

Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is
x
2 / 4 - y 2 / 9 = 1

Problem 2

Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is
16 y
2 - x 2 = 16

Problem 3

Find the equation of a hyperbola that has the y axis as the transverse axis, a center at (0 , 0) and passes through the points (0 , 5) and (2 , 5√2).

Problem 4

Find the equation of a hyperbola whose vertices are at (0 , -3) and (0 , 3) and has a focus at (0 , 5).

Problem 5

Find the asymptotes of the parabolas given by the equations:
a) x
2 / 4 - y 2 / 36 = 1
b) y
2 - 49 x 2 = 49

Problem 6

Find the equation of a hyperbola with vertices at (0 , -7) and (0 , 7) and asymptotes given by the equations y = 3x and y = - 3x.

Problem 7

Find the equation of a hyperbola with foci at (-2 , 0) and (2 , 0) and asymptotes given by the equation y = x and y = -x.

Problem 8

Write the equation of a hyperbola with foci at (-1 , 0) and (1 , 0) and one of its asymptotes passes through the point (1 , 3).

Problem 9

Write the equation of a hyperbola with the x axis as its transverse axis, point (3 , 1) lies on the graph of this hyperbola and point (4 , 2) lies on the asymptote of this hyperbola.

Problem 10

Find the equation of each parabola shown below. The graphs in b) and c) also shows the asymptotes.
a)
b) c)

Solution to Problem 1

Transverse axis: x axis or y = 0
center at (0 , 0)
vertices at (2 , 0) and (-2 , 0)
c
2 = 4 + 9 = 13. Foci are at (√13 , 0) and (-√13 , 0).

Solution to Problem 2

Divide all terms of the given equation by 16 which becomes y 2 - x 2 / 16 = 1
Transverse axis: y axis or x = 0
center at (0 , 0)
vertices at (0 , 1) and (0 , -1)
c
2 = 1 + 16 = 17. Foci are at (0 , √17) and (0 , -√17).

Solution to Problem 3

Since the y axis is the transverse axis, the equation has the form y 2 / a 2 - x 2 / b 2 = 1
Use the point (0 , 5) to write: (5)
2 / a 2 = 1 and find a 2 = 25. Use the second point to write (5√2) 2 / 25 - 2 2 / b 2 = 1 and find b 2 = 4
The equation is given by: y
2 / 25 - x 2 / 4 = 1

Solution to Problem 4

Since the vertices are at (0,-3) and (0,3), the transverse axis is the y axis and the center is at (0,0). The equation has the form: y 2 / 9 - x 2 / b 2 = 1, a 2 = 9.
The focus is at (0,5) hence c = 5. We now use the formula c
2 = a 2 + b 2 to find b 2 = 25 - 9 = 16
The equation may be written as: y
2 / 9 - x 2 / 16 = 1

Solution to Problem 5

a) y = 3x and y = -3x
b) y = 7x and y = -7x

Solution to Problem 6

Since the vertices are at (0,-7) and (0,7), the transverse axis of the hyperbola is the y axis, the center is at (0,0) and the equation of the hyperbola ha s the form y 2 / a 2 - x 2 / b 2 = 1 with a 2 = 49. The asymptote is given by y = +or-(a/b)x, hence a/b = 3 which gives a 2 = 9 b 2 .
Since a
2 = 49, 9 b 2 = 49 and b 2 = 49/9
The equation of the hyperbola is given by: y
2 / 49 - 9 x 2 / 49 = 1

Solution to Problem 7

Since the foci are at (-2,0) and (2,0), the transverse axis of the hyperbola is the x axis, the center is at (0,0) and the equation of the hyperbola has the form x 2 / a 2 - y 2 / b 2 = 1 with c 2 = 4 = a 2 + b 2
The asymptote is given by y = +or-(b/a)x, hence a/b = 1 which gives a
2 = b 2 .
Solve the two equations 4 = a
2 + b 2 and a 2 = b 2 to find: a 2 = 2 and b 2 = 2.
The equation of the hyperbola is give by: x
2 / 2 - y 2 / 2 = 1

Solution to Problem 8

Since the foci are at (-1,0) and (1,0), the transverse axis of the hyperbola is the x axis, the center is at (0,0) and the equation of the hyperbola has the form x 2 / a 2 - y 2 / b 2 = 1 with c 2 = 1 2 = a 2 + b 2
The asymptote is given by y = (b/a)x, hence a/b = 3/1 = 3 which gives a
2 = 9 b 2 .
Solve both equations: 1 = a
2 + b 2 and a 2 = 9 b 2 .
Solve to find: b
2 = 1/10 and a 2 = 9/10
The equation of the hyperbola is given by: (10/9) x
2 / - 10 y 2 / b 2 = 1

Solution to Problem 9

The equation of the hyperbola has the form: x 2 / a 2 - y 2 / b 2 = 1
Use point (3 , 1) to write: 3
2 / a 2 - 1 2 / b 2 = 1
The asymptote has the form: y = + or - (b/a)x, using the point (4,2) that lies on the asymptote we write: b / a = 2/4 = 1/2 or 4 b
2 = a 2
Solve the two equations to find: a
2 = 5 and b 2 = 5/4
The equation of the hyperbola has the form: x
2 / 5 - y 2 / (5/4) = 1

Solution to Problem 10

a)
Vertices at (-1,0) and (1,0) and point (-3,2) lies on the hyperbola.
Equation: x
2 - y 2 / 0.5 = 1
b)
Vertices at (-2,0) and (2,0) and point (2,2) lies on one asymptote.
Equation: x
2 / 4 - y 2 / 4 = 1
c)
Vertices at (0 , 0.5) and (0 , -0.5) and asymptote y = x/6.
Equation: y
2 / 0.25 - x 2 / 9 = 1