Question

In Problems 19–28, find an equation for the hyperbola described. Graph the equation by hand. Vertices at $(-4,0)$and $(4,0)$; asymptote the line$y=2x$.

Short answer

Equation of the hyperbola:$\frac{x^2}{16}-\frac{y^2}{64}=1$

Graph of the hyperbola:

Step-by-step solution

Step 1. Given information 

Hyperbola with vertices at $(-4,0)$and role="math" localid="1646997921363" $(4,0)$; asymptote the line$y=2x$.

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 vertices at $(-4,0)$and $(4,0)$; asymptote the line $y=2x$.

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

We get,

$$\begin{gathered} a=4,\frac{b}{a}=2 \\ \Rightarrow b=2a \\ \Rightarrow b=2\times 4 \\ \Rightarrow b=8 \end{gathered}$$

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

$\frac{x^2}{16}-\frac{y^2}{64}=1$

Step 4. Graphing the hyperbola 

Graph of the hyperbola: