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.$4y^2-x^2=16$

Short answer

$$\begin{gathered} \mathrm{Center}:(0,0) \\ \mathrm{Transverse}\mathrm{axis}:y-axis \\ \mathrm{Vertices}:(0,\pm 2) \\ \mathrm{Foci}:(0,\pm 2\sqrt{5}) \\ \mathrm{Asymptotes}:y=\pm \frac{1}{2}x \end{gathered}$$

Graph of the equation:

Step-by-step solution

Step 1. Given information 

Equation of hyperbola, $4y^2-x^2=16$

Writing the equation in standard form,

$$\begin{gathered} \frac{4y^2}{16}-\frac{x^2}{16}=1 \\ \Rightarrow \frac{y^2}{4}-\frac{x^2}{16}=1 \end{gathered}$$

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,

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

$$\begin{gathered} \mathrm{Center}:(0,0) \\ \mathrm{Transverse}\mathrm{axis}:y-axis \\ \mathrm{Vertices}:(0,\pm \mathrm{a})=(0,\pm 2) \\ \mathrm{Foci}:(0,\pm \mathrm{c})=(0,\pm 2\sqrt{5}) \\ \mathrm{Asymptotes}:y=\pm \frac{a}{b}x \\ \Rightarrow y=\pm \frac{2}{4}x \\ \Rightarrow y=\pm \frac{1}{2}x \end{gathered}$$

Step 3. Graph of the equation

Graph of the equation is: