Chapter 9.5, Problem 18PPS

To determine

To Plot: The graph of given hyperbola

Explanation of Solution

Given:

  $\frac{{\left(x-3\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{4}=1$

Formula Used:

  $\begin{array}{l}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\\ y=\frac{b}{a}xor=-\frac{b}{a}x\\ e=\frac{\sqrt{a^2+b^2}}{a}\\ l=\frac{2b^2}{a}\end{array}$

Calculation:

  

  $\begin{array}{l}a=4\\ b=2\\ \left(h,k\right)=\left(3,-1\right)\\ p=\frac{b^2}{\sqrt{a^2+b^2}}=\frac{4}{\sqrt{4+16}}\\ =\frac{4}{4\sqrt{5}}=\frac{1}{\sqrt{5}}\\ \text{Asymptotes}\\ y=\pm \frac{2}{4}\left(x-3\right)+6=\pm \frac{1}{2}\left(x-3\right)+6\\ Vertices\\ \left(-1,-1\right)\&\left(7,-1\right)\end{array}$