# 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 isx

^{ 2}/ 4 - y

^{ 2}/ 9 = 1

## Problem 2

Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is16 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)

## Solutions to the Above Problems

## Solution to Problem 1

Transverse axis: x axis or y = 0center 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 xb) 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

### More References and links

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