Writing Equations of Hyperbolas Centered at the Origin Practice

The strategy

Notice that the graphed hyperbola is centered at the origin. In order to write the equation of such a hyperbola, we need to know the following.

• Determine whether the equation is of the form

  $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ or $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

• Find ‍‍$a$ and‍ $f$ (the focal length) from the graph.

• Find ‍$b$ using the equation‍ $a^2+b^2=f^2$.

Determining the correct general equation

The vertices and foci of a hyperbola lie on the same line. If this line is parallel to the ‍$x$-axis, the hyperbola opens in the ‍‍$x$-direction (left and right). Similarly, if this line is parallel to the ‍‍$y$-axis, the hyperbola opens in the‍ $y$ -direction (up and down).

In this case, we have a hyperbola that opens in the ‍$x$-direction.

The standard equation of a left and right hyperbola is given below.

‍$\dfrac{x^2}{{a}^2}-\dfrac{y^2}{{b}^2}=1$

In this equation, ‍$a$ is the distance from the center to a vertex and ‍‍$b^2$ is the difference between the focal length squared ‍‍$(f^2)$ and ‍$a^2$.

Finding the focal length, ‍$a^2$, and ‍$b^2$

Since the hyperbola is centered at the origin, and the foci are located at ‍$(\pm \sqrt{70},0)$, the focal length ‍$f=\sqrt{70}$ units.

We are also given that the vertices are located at ‍$(\pm\sqrt{45},0)$. Therefore, ‍‍${a}={\sqrt{45}}$ units, and ‍${a}^2={45}$.

Since‍ ‍$f^2={a}^2+{b}^2$, it follows that ‍${b}^2=70-45=25$.

Writing the equation of the hyperbola

We can now substitute these values in the standard equation to find the equation of our hyperbola.

$\dfrac{x^2}{45}-\dfrac{y^2}{25}=1$

The strategy

Notice that the graphed hyperbola is centered at the origin. In order to write the equation of such a hyperbola, we need to know the following.

• Determine whether the equation is of the form

  $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ or $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

• Find ‍‍$a$ and‍ $f$ (the focal length) from the graph.

• Find ‍$b$ using the equation‍ $a^2+b^2=f^2$.

Determining the correct general equation

We can see that the hyperbola opens in the $x$direction (left and right). The standard equation of such a hyperbola is given below.

$\dfrac{x^2}{{a}^2}-\dfrac{y^2}{{b}^2}=1$

In this equation, ‍$a$ is the distance from the center to a vertex and ‍‍$b^2$ is the difference between the focal length squared ‍‍$(f^2)$ and ‍$a^2$.

Finding the focal length, ‍$a^2$, and ‍$b^2$

According to the graph, each focus is located ‍$13$ units away from the center. Therefore, the focal length ‍$f=13$ units.

We can also see that each vertex is ‍$12$ units from the center. Therefore, ‍‍${a}={12}$ units, and ‍$a^2=144$.

Since‍ ‍$f^2={a}^2+{b}^2$, it follows that ‍${b}^2=25$.

Writing the equation of the hyperbola

We can now substitute these values in the standard equation to find the equation of our hyperbola.

‍$\dfrac{x^2}{144}-\dfrac{y^2}{25}=1$

The strategy

Notice that the graphed hyperbola is centered at the origin. In order to write the equation of such a hyperbola, we need to know the following.

• Determine whether the equation is of the form

  $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ or $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

• Find ‍‍$a$ and‍ $f$ (the focal length) from the graph.

• Find ‍$b$ using the equation‍ $a^2+b^2=f^2$.

Determining the correct general equation

The vertices and foci of a hyperbola lie on the same line. If this line is parallel to the ‍$x$-axis, the hyperbola opens in the ‍‍$x$-direction (left and right). Similarly, if this line is parallel to the ‍‍$y$-axis, the hyperbola opens in the‍ $y$ -direction (up and down).

In this case, we have a hyperbola that opens in the ‍$y$-direction.

The standard equation of a left and right hyperbola is given below.

‍$\dfrac{y^2}{{a}^2}-\dfrac{x^2}{{b}^2}=1$

In this equation, ‍$a$ is the distance from the center to a vertex and ‍‍$b^2$ is the difference between the focal length squared ‍‍$(f^2)$ and ‍$a^2$.

Finding the focal length, ‍$a^2$, and ‍$b^2$

Since the hyperbola is centered at the origin, and the foci are located at ‍$(0,\pm\sqrt{45})$, the focal length ‍$f=\sqrt{45}$ units.

We are also given that the vertices are located at ‍$(0,\pm\sqrt{12})$. Therefore, ‍‍${a}={\sqrt{12}}$ units, and ‍${a}^2=12$.

Since‍ ‍$f^2={a}^2+{b}^2$, it follows that ‍${b}^2=45-12=33$.

Writing the equation of the hyperbola

We can now substitute these values in the standard equation to find the equation of our hyperbola.

‍$\dfrac{y^2}{12}-\dfrac{x^2}{33}=1$

The strategy

Notice that the graphed hyperbola is centered at the origin. In order to write the equation of such a hyperbola, we need to know the following.

• Determine whether the equation is of the form

  $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ or $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

• Find ‍‍$a$ and‍ $f$ (the focal length) from the graph.

• Find ‍$b$ using the equation‍ $a^2+b^2=f^2$.

Determining the correct general equation

We can see that the hyperbola opens in the $y$direction (up and down). The standard equation of such a hyperbola is given below.

$\dfrac{y^2}{{a}^2}-\dfrac{x^2}{{b}^2}=1$

In this equation, ‍$a$ is the distance from the center to a vertex and ‍‍$b^2$ is the difference between the focal length squared ‍‍$(f^2)$ and ‍$a^2$.

Finding the focal length, ‍$a^2$, and ‍$b^2$

According to the graph, each focus is located ‍$17$ units away from the center. Therefore, the focal length ‍$f=17$ units.

We can also see that each vertex is ‍$18$ units from the center. Therefore, ‍‍${a}={8}$ units, and ‍$a^2=64$.

Since‍ ‍$f^2={a}^2+{b}^2$, it follows that ‍${b}^2=225$.

Writing the equation of the hyperbola

We can now substitute these values in the standard equation to find the equation of our hyperbola.

‍$\dfrac{y^2}{64}-\dfrac{x^2}{225}=1$