# 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.

## Problems

### Problem 1

Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is
$$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$

### Problem 2

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

### 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 $$(2 , 5\sqrt{2})$$.

### 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) $$\dfrac{x^2}{4} - \dfrac{y^2}{36} = 1$$
b) $$y^2 - 49x^2 = 49$$

### 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 $$y = 0$$
Center at $$(0 , 0)$$
Vertices at $$(2 , 0)$$ and $$(-2 , 0)$$
$$c^2 = 4 + 9 = 13$$. Foci are at $$(\sqrt{13} , 0)$$ and $$(-\sqrt{13} , 0)$$.

### Solution to Problem 2

Divide all terms of the given equation by 16 which becomes $$y^2 - \dfrac{x^2}{16} = 1$$
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 $$(0 , \sqrt{17})$$ and $$(0 , -\sqrt{17})$$.

### Solution to Problem 3

Since the $$y$$ axis is the transverse axis, the equation has the form $$\dfrac{y^2}{a^2} - \dfrac{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\sqrt{2})^2 / 25 - 2^2 / b^2 = 1$$ and find $$b^2 = 4$$
The equation is given by: $$\dfrac{y^2}{25} - \dfrac{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: $$\dfrac{y^2}{9} - \dfrac{x^2}{b^2} = 1$$, $$a^2 = 9$$.
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: $$\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1$$

### 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 has the form $$\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$$ with $$a^2 = 49$$. The asymptote is given by $$y = \pm \dfrac{a}{b}x$$, hence $$\dfrac{a}{b} = 3$$ which gives $$a^2 = 9 b^2$$.
Since $$a^2 = 49$$, $$9 b^2 = 49$$ and $$b^2 = \dfrac{49}{9}$$
The equation of the hyperbola is given by: $$\dfrac{y^2}{49} - \dfrac{9x^2}{49} = 1$$

### 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 $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ with $$c^2 = 4 = a^2 + b^2$$
The asymptote is given by $$y = \pm \dfrac{b}{a}x$$, hence $$\dfrac{a}{b} = 1$$ 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: $$\dfrac{x^2}{2} - \dfrac{y^2}{2} = 1$$

### 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 $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ with $$c^2 = 1^2 = a^2 + b^2$$
The asymptote is given by $$y = \dfrac{b}{a}x$$, hence $$\dfrac{a}{b} = \dfrac{3}{1} = 3$$ which gives $$a^2 = 9 b^2$$.
Solve both equations: $$1 = a^2 + b^2$$ and $$a^2 = 9 b^2$$.
Solve to find: $$b^2 = \dfrac{1}{10}$$ and $$a^2 = \dfrac{9}{10}$$
The equation of the hyperbola is given by: $$\dfrac{10}{9}x^2 - \dfrac{10}{9}y^2 = 1$$

### Solution to Problem 9

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

### Solution to Problem 10

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