Chapter A5.2, Problem 24E

To determine

To graph: thehyperbola and explain what happens to the graph if eccentricity increases.

Answer to Problem 24E

As the value of $e$ increases, the slope increases.

The equation of hyperbola can now be written as: $\frac{x^2}{a^2}-\frac{y^2}{a^2\left(e^2-1\right)}=1$

Explanation of Solution

Given:

The general equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

Calculation:

Therelations between $c,e,a$ and $b$ for the general equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{a^2\left(e^2-1\right)}=1$

That is,

  $b^2=c^2-a^2$ And $c=ae$

This implies that,

  $\begin{array}{l}{b}^2=a^2{e}^2-a^2\\ \text{ =}{a}^2\left(e^2-1\right)\end{array}$

Hence the equation of hyperbola can now be written as:

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

The asymptotes of this hyperbola thus, have the general equation as:

  $y=\left(\pm \sqrt{e^2-1}\right)x$

As the value of $e$ increases, the slope increases. If the value of $e$ is increased to a very high value,

Then the hyperbola approaches to a single vertical line, which is the y-axis.

To draw a graph, let us assume $a=1$ ,and let us take various values for $e=(2,3,4,5)$ .so we

Draw the graphs for:

  $\begin{array}{l}{x}^2-\frac{y^2}{3}=1\\ x^2-\frac{y^2}{8}=1\\ x^2-\frac{y^2}{15}=1\\ x^2-\frac{y^2}{24}=1\end{array}$

  $x^2-\frac{y^2}{3}=1$

  

  $x^2-\frac{y^2}{8}=1$

  

  $x^2-\frac{y^2}{15}=1$

  

  $x^2-\frac{y^2}{24}=1$

  

Conclusion:

Therefore, the equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{a^2\left(e^2-1\right)}=1$

As the value of $e$ increases, the slope increases.