Chapter 8.6, Problem 5E

Solution

Graph the given equation and identify the important characteristics of graph.

The graph of the given equation is

  

With centre at $\left(6,-1\right)$ and vertices at $\left(11,-1\right),\left(1,-1\right)$

Given information:

Given equation is:

  $\frac{{\left(x-6\right)}^2}{25}-{\left(y+1\right)}^2=1$

Formula used:

Standard form of equation of a hyperbola with center $\left(h,k\right)$ and vertices $\left(h\pm a,k\right)$ is

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

Calculation:

The given equation can be rewritten as

  $\frac{{\left(x-6\right)}^2}{5^2}-\frac{{\left(y-\left(-1\right)\right)}^2}{1^2}=1$

Now in order to graph the given equation

  $\frac{{\left(x-6\right)}^2}{5^2}-\frac{{\left(y-\left(-1\right)\right)}^2}{1^2}=1$ ,

First identify its center and vertices by comparing the equation with standard form of the equation of the hyperbola:

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , with centre $\left(h,k\right)$ .

Comparing above with $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ gives

  $\left(h,k\right)=\left(6,-1\right)$ and vertices $\left(6\pm 5,-1\right)=\left(11,-1\right),\left(1,-1\right)$

Thus, the given hyperbola has centre at $\left(6,-1\right)$ and vertices at $\left(11,-1\right),\left(1,-1\right)$

Thus, the graph of this hyperbola will be as

  

With centre at $\left(6,-1\right)$ and vertices at $\left(11,-1\right),\left(1,-1\right)$