Writing Equations of Hyperbolas Centered at the Origin Practice

1. A hyperbola centered at the origin has vertices at ‍$(\pm \sqrt{45},0)$ and foci at ‍$(\pm \sqrt{70},0)$.

   Write the equation of this hyperbola.

2. Write the equation of the hyperbola graphed below, whose vertices and foci are marked.

   
   

3. A hyperbola centered at the origin has vertices at ‍$(0,\pm \sqrt{12})$ and foci at ‍$(0,\pm \sqrt{45})$.

   Write the equation of this hyperbola.

4. Write the equation of the hyperbola graphed below, whose vertices and foci are marked.

   
   

Source: Khan Academy, https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:hyperb-foci/e/equation_of_a_hyperbola
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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 ‍$x$-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.

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

   ---

2. 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.

   ‍$\dfrac{x^2}{144}-\dfrac{y^2}{25}=1$

   ---

3. 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.

   ‍$\dfrac{y^2}{12}-\dfrac{x^2}{33}=1$

   ---

4. 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.

   ‍$\dfrac{y^2}{64}-\dfrac{x^2}{225}=1$