Question

In Problems 29–36, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.$\frac{y^2}{16}-\frac{x^2}{4}=1$

Short answer

$$\begin{gathered} \mathrm{Center}:(0,0) \\ \mathrm{Transverse}\mathrm{axis}:y-axis \\ \mathrm{Vertices}:(0,\pm 4) \\ \mathrm{Foci}:(0,\pm 2\sqrt{5}) \\ \mathrm{Asymptotes}:y=\pm 2x \end{gathered}$$

Graph of the equation is:

Step-by-step solution

Step 1. Given information 

Equation of hyperbola,$\frac{y^2}{16}-\frac{x^2}{4}=1$

Step 2. Finding the center, transverse axis, vertices, foci, and asymptotes of hyperbola

Comparing the given equation with standard form, $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,b^2=c^2-a^2$

We get,

role="math" localid="1647141861700" $$\begin{gathered} a=4,b=2, \\ c=\pm \sqrt{a^2+b^2} \\ \Rightarrow c=\pm \sqrt{4^2+2^2} \\ \Rightarrow c=\pm \sqrt{16+4} \\ \Rightarrow c=\pm \sqrt{20} \\ \Rightarrow c=\pm 2\sqrt{5} \end{gathered}$$

$$\begin{gathered} \mathrm{Center}:(0,0) \\ \mathrm{Transverse}\mathrm{axis}:y-axis \\ \mathrm{Vertices}:(0,\pm \mathrm{a})=(0,\pm 4) \\ \mathrm{Foci}:(0,\pm \mathrm{c})=(0,\pm 2\sqrt{5}) \\ \mathrm{Asymptotes}:y=\pm \frac{a}{b}x \\ \Rightarrow y=\pm \frac{4}{2}x \\ \Rightarrow y=\pm 2x \end{gathered}$$

Step 3. Graph of the equation

Graph of the equation is: