Question

In Problems 19–28, find an equation for the hyperbola described. Graph the equation by hand. Center at $(0,0)$; focus at $(0,-6)$; vertex at$(0,4)$.

Short answer

Equation of the hyperbola is:$\frac{y^2}{16}-\frac{x^2}{20}=1$

Graph of the hyperbola is:

Step-by-step solution

Step 1. Given information  

Hyperbola withCenter at $(0,0)$; focus at $(0,-6)$; vertex at $(0,4)$.

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 $(0,-6)$; vertex at $(0,4)$.

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

We get, $a=4,c=6$

$$\begin{gathered} b^2=c^2-a^2 \\ \Rightarrow b^2=6^2-4^2 \\ \Rightarrow b^2=36-16 \\ \Rightarrow b^2=20 \\ \Rightarrow b=\pm 2\sqrt{5} \end{gathered}$$

Substituting $a$and $b$in the equation,

role="math" localid="1646820641211" $\frac{y^2}{16}-\frac{x^2}{20}=1$

Step 4. Graphing the hyperbola

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

$$\begin{gathered} \frac{{\left(\pm 6\right)}^2}{16}-\frac{x^2}{20}=1 \\ \Rightarrow \frac{x^2}{20}=\frac{36}{16}-1 \\ \Rightarrow x^2=\frac{20\times (36-16)}{16} \\ \Rightarrow x=\pm 5 \end{gathered}$$

The graph of the hyperbola is: