Chapter 4.2, Problem 66E

(a)

To determine

To prove: That the equation of hyperbola can be rearranged into given way.

Explanation of Solution

Given information: The equation of hyperbola is given as,

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Calculation: Rearranging the equation,

  $\begin{array}{l}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{y^2}{b^2}=\frac{x^2}{a^2}-1\\ y^2=b^2\times \frac{\left(x^2-a^2\right)}{a^2}\\ y=\pm \frac{b}{a}\times \sqrt{\left(x^2-a^2\right)}\end{array}$

Hence, it is proved that the equation of hyperbola can be transformed into another form.

(b)

To determine

To prove: That $g\left(x\right)=\left(\frac{b}{a}\right)\left|x\right|$ is an end behavior model for the given equation.

Explanation of Solution

Given information: The equation is given as,

  $f\left(x\right)=\left(\frac{b}{a}\right)\sqrt{\left(x^2-a^2\right)}$

End behavior model describes the behavior of function as x approaches to infinity.

Proof:Calculatethe value of function at x approaches infinity,

  $\begin{array}{l}g\left(x\right)=\underset{x\to \infty }{\lim }f\left(x\right)\\ g\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{b}{a}\right)\times \sqrt{\left(x^2-a^2\right)}\\ \text{as}x>>>a\text{so}\text{eliminate}a,\\ g\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{b}{a}\right)\times \sqrt{x^2}\\ \text{as}\text{root}\text{value}\text{cannot}\text{be}\text{negative}\text{so}\text{mod}\text{is}\text{taken,}\\ g\left(x\right)=\left(\frac{b}{a}\right)\times \left|x\right|\end{array}$

Hence, it is proved that $g\left(x\right)=\left(\frac{b}{a}\right)\times \left|x\right|$ is an end behavior model for the given equation.

(c)

To determine

To prove: That $g\left(x\right)=-\left(\frac{b}{a}\right)\left|x\right|$ is an end behavior model for the given equation.

Explanation of Solution

Given information: The equation is given as,

  $f\left(x\right)=-\left(\frac{b}{a}\right)\sqrt{\left(x^2-a^2\right)}$

End behavior model describes the behavior of function as x approaches to infinity.

Proof: Calculatethe value of function at x approaches infinity,

  $\begin{array}{l}g\left(x\right)=\underset{x\to \infty }{\lim }f\left(x\right)\\ g\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{-b}{a}\right)\times \sqrt{\left(x^2-a^2\right)}\\ \text{as}x>>>a\text{so}\text{eliminate}a,\\ g\left(x\right)=\underset{x\to \infty }{\lim }\left(\frac{-b}{a}\right)\times \sqrt{x^2}\\ \text{as}\text{root}\text{value}\text{cannot}\text{be}\text{negative}\text{so}\text{mod}\text{is}\text{taken,}\\ g\left(x\right)=-\left(\frac{b}{a}\right)\times \left|x\right|\end{array}$

Hence, it is proved that $g\left(x\right)=-\left(\frac{b}{a}\right)\times \left|x\right|$ is an end behavior model for the given equation.