# 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)

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

## 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 = ~+mn~ (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 = ~+mn~ (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 = ~+mn~ (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