Question

In Problems 41– 48, find an equation for the hyperbola described. Graph the equation by hand. Center at $(-3,-4)$; focus at $(-3,-8)$; vertex atrole="math" localid="1647151361836" $(-3,-2)$.

Short answer

Equation of the hyperbola is:$\frac{{(y+4)}^2}{4}-\frac{{(x+3)}^2}{12}=1$

Graph of the hyperbola is:

Step-by-step solution

Step 1. Given information 

Hyperbola with center at $(-3,-4)$; focus at $(-3,-8)$; vertex at$(-3,-2)$.

Step 2. Finding the equation of the hyperbola  

Comparing the given hyperbola with the standard form,

$$\begin{gathered} \frac{{(y-h)}^2}{a^2}-\frac{{(x-k)}^2}{b^2}=1,b^2=c^2-a^2 \\ \mathrm{Center}:(h,k)=(-3,-4) \\ \mathrm{Focus}:(h,-c+k)=(-3,-8) \\ \Rightarrow c=4 \\ \mathrm{Vertex}:(h,a+k)=(-3,-2) \\ \Rightarrow a=2 \\ {b}^2=c^2-a^2 \\ \Rightarrow b^2=4^2-2^2 \\ \Rightarrow b^2=16-4 \\ \Rightarrow b^2=12 \\ \Rightarrow b=\pm 2\sqrt{3} \end{gathered}$$

Substituting for $h,k,a\mathrm{and}b$in the standard equation,

$\frac{{(y+4)}^2}{4}-\frac{{(x+3)}^2}{12}=1$

Step 3. Graph of the hyperbola 

Equations of asymptotes:

$$\begin{gathered} (y-k)=\pm \frac{a}{b}(x-h) \\ \Rightarrow (y+4)=\pm \frac{2}{2\sqrt{3}}(x+3) \\ \Rightarrow (y+4)=\pm \frac{1}{\sqrt{3}}(x+3) \end{gathered}$$

Graph of the hyperbola is: