Question

In Problems 41– 48, find an equation for the hyperbola described. Graph the equation by hand. Center at $(1,4)$; focus at $(-2,4)$; vertex at$(0,4)$.

Short answer

Equation of the hyperbola is:${(x-1)}^2-\frac{{(y-4)}^2}{8}=1$

Graph of the hyperbola is:

Step-by-step solution

Step 1. Given information 

Hyperbola with center at $(1,4)$; focus at $(-2,4)$; vertex at $(0,4)$.

Step 2. Finding the equation of the hyperbola  

Comparing the given hyperbola with the standard form,

$$\begin{gathered} \frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1,b^2=c^2-a^2 \\ \mathrm{Center}:(h,k)=(1,4) \\ \mathrm{Focus}:(-c+h,k)=(-2,4) \\ \Rightarrow c=3 \\ \mathrm{Vertex}:(-a+h,k)=(0,4) \\ \Rightarrow a=1 \\ {b}^2=c^2-a^2 \\ \Rightarrow b^2=3^2-1^2 \\ \Rightarrow b^2=9-1 \\ \Rightarrow b^2=8 \\ \Rightarrow b=\pm 2\sqrt{2} \end{gathered}$$

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

$$\begin{gathered} \frac{{(x-1)}^2}{1}-\frac{{(y-4)}^2}{8}=1 \\ \Rightarrow {(x-1)}^2-\frac{{(y-4)}^2}{8}=1 \end{gathered}$$

Step 3. Graph of the hyperbola 

Equations of asymptotes:

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

Graph of the hyperbola is: