Question

In Problems 19–28, find an equation for the hyperbola described. Graph the equation by hand. Vertices at $(0,-6)$and $(0,6)$; asymptote the line$y=2x$.

Short answer

Equation of the hyperbola:$\frac{y^2}{36}-\frac{x^2}{9}=1$

Graph of the hyperbola:

Step-by-step solution

Step 1. Given information 

Hyperbola with vertices at $(0,-6)$and $(0,6)$; 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 $(0,-6)$and $(0,6)$; asymptote the line $y=2x$.

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

We get,

role="math" localid="1646999569672" $$\begin{gathered} a=6,\frac{a}{b}=2 \\ \Rightarrow b=\frac{a}{2} \\ \Rightarrow b=\frac{6}{2} \\ \Rightarrow b=3 \end{gathered}$$

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

$\frac{y^2}{36}-\frac{x^2}{9}=1$

Step 4. Graphing the hyperbola 

Graph of the hyperbola: