Question

In Problems 19–28, find an equation for the hyperbola described. Graph the equation by hand. Center at $(0,0)$; focus at $(0,5)$; vertex at $(0,3)$.

Short answer

Equation of the hyperbola is:$\frac{y^2}{9}-\frac{x^2}{16}=1$

Graph of the hyperbola is:

Step-by-step solution

Step 1. Given information  

Hyperbola with Center at $(0,0)$; focus at $(0,5)$; vertex at$(0,3)$.

Step 2. Required to find 

Find the equation of the hyperbola. Graph the equation.

Step 3. Finding the equation of the hyperbola 

Given, Hyperbola with Center at $(0,0)$; focus at $(0,5)$; vertex at $(0,3)$.

Comparing with standard form of hyperbola, localid="1646825525683" $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,b^2=c^2-a^2;\mathrm{foci}=(0,\pm c);\mathrm{vertices}=(0,\pm a)$

We get, localid="1646825530337" $a=3,c=5$

$$\begin{gathered} b^2=c^2-a^2 \\ \Rightarrow b^2=5^2-3^2 \\ \Rightarrow b^2=25-9 \\ \Rightarrow b^2=16 \\ \Rightarrow b=\pm 4 \end{gathered}$$

Substituting $a$and role="math" localid="1646825552217" $b$in the equation,

localid="1646825556348" $\frac{y^2}{9}-\frac{x^2}{16}=1$

Step 4. Graphing the hyperbola

To determine the opening of hyperbola, substitute $x=\pm 4$in the equation and find $y$.

$$\begin{gathered} \frac{y^2}{9}-\frac{{\left(\pm 4\right)}^2}{16}=1 \\ \Rightarrow \frac{y^2}{9}=1+1 \\ \Rightarrow y^2=18 \\ \Rightarrow y=\pm 3\sqrt{2} \end{gathered}$$

The graph of the hyperbola is: