Question

In Problems 19–28, find an equation for the hyperbola described. Graph the equation by hand. Center at $(0,0)$; focus at $(-3,0)$; vertex at$(2,0)$.

Short answer

Equation of the hyperbola is:$\frac{x^2}{4}-\frac{y^2}{5}=1$

Graph of the hyperbola is:

Step-by-step solution

Step 1. Given information  

Hyperbola withCenter at $(0,0)$; focus at $(-3,0)$; vertex at $(2,0)$.

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 $(-3,0)$; vertex at $(2,0)$.

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

We get, $a=2,c=3$

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

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

$\frac{x^2}{4}-\frac{y^2}{5}=1$

Step 4. Graphing the hyperbola

To determine the opening of hyperbola, substitute $x=\pm 3$in the equation and find $y$.

$$\begin{gathered} \frac{{\left(\pm 3\right)}^2}{4}-\frac{y^2}{5}=1 \\ \Rightarrow \frac{y^2}{5}=\frac{9}{4}-1 \\ \Rightarrow y^2=\frac{5\times 5}{4} \\ \Rightarrow y=\pm \frac{5}{2} \end{gathered}$$

The graph of the hyperbola is: