Chapter 8.3, Problem 69E

Solution

a.

To graph: The coordinate axes in the construction window.

Given information:

Use a geometry software package, such as Cabri Geometry $I{I}^{TM}$ The Geometer’s Sketchpad®.

Graph:

  

Interpretation:

The coordinated axes have been plotted using graphing utility.

b.

To graph: To construct the two points on the x-axis at $\left(\pm 5,0\right)$ as the foci.

Given information:

The coordinates $\left(\pm 5,0\right)$ .

Graph:

  

Interpretation:

The two points on the x-axis at $\left(\pm 5,0\right)$ as the foci has been constructed in the above graph.

c.

To graph: To construct concentric circles of radii $r=1,2,3,\mathrm{.....},12$ centered at these two foci.

Given information:

Concentric circles of radii $r=1,2,3,\mathrm{.....},12$ centered at these two foci.

Graph:

  

Interpretation:

The concentric circles of radii $r=1,2,3,\mathrm{.....},12$ centered at these two foci have been constructed in the above graph.

d.

To graph: To construct the points where these concentric circles meet and have a

diference of radii of $2a=6$ .

Given information:

The distance between the vertices is $2a$ .

Graph:

  

Interpretation:

Start with the circle's radius of $12$ to get a sense of where the points of intersection are where the difference in radii is at six.

Thus to locate the circle a with radius $6$ their difference to be $6$ .

  $11/5,10/4,9/3\text{ and }8/2$ are the other combinations.

Therefore, the points where these concentric circles meet and have a difference of radii of $2a=6$ has been constructed in the above graph.

e.

To find: The equation whose graph includes all of these points.

The equation whose graph includes all of these points is $\frac{x^2}{9}-\frac{y^2}{16}=1$ .

Given information:

The distance between the vertices is $2a$ .

Calculation:

The standard form of the hyperbola is given by,

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The distance between the vertices is $2a$ .

  $\begin{array}{c}2a=6\\ a=3\end{array}$

Let’s consider the distance between the foci is $2c$ .

  $\begin{array}{c}2c=10\\ c=5\end{array}$

By using the Pythagorean Theorem,

  $\begin{array}{l}{c}^2=a^2+b^2\\ b^2=c^2-a^2\\ b^2=5^2-3^2\\ b^2=16\end{array}$

Substituting the values in the above equation to get,

  $\begin{array}{c}\frac{x^2}{3^2}-\frac{y^2}{16}=1\\ \frac{x^2}{9}-\frac{y^2}{16}=1\end{array}$

Therefore, the equation whose graph includes all of these points is $\frac{x^2}{9}-\frac{y^2}{16}=1$ .