Chapter A5.3, Problem 41E

(a)

To determine

To find: The effect of rotating ${180}^{\circ }$ on the given equation.

Answer to Problem 41E

Here, the equation for the ellipse is $\frac{x{\text{'}}^2}{b^2}+\frac{y{\text{'}}^2}{a^2}=1$ .

Explanation of Solution

Given: $\alpha ={90}^{\circ }$ .

Calculation:

Given,

  $\alpha ={90}^{\circ }$

Therefore,

  $\begin{array}{l}\sin \alpha =1\\ \cos \alpha =0\end{array}$

Now, after rotation of axes,

  $\begin{array}{l}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ \text{ =}-y\text{'}\\ y=x\text{'}\sin \alpha -y\text{'}\cos \alpha \\ \text{ =}x\text{'}\end{array}$

Given equation is:

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Substituting the values of $x=-y\text{'}$ and $y=x\text{'}$ , we have

  $\frac{{\left(-y\text{'}\right)}^2}{a^2}+\frac{{\left(x\text{'}\right)}^2}{b^2}=1$

  $\Rightarrow \frac{x{\text{'}}^2}{b^2}+\frac{y{\text{'}}^2}{a^2}=1$

Conclusion:

Hence,the equation for the ellipse is $\frac{x{\text{'}}^2}{b^2}+\frac{y{\text{'}}^2}{a^2}=1$ .

(b)

To determine

To find: The effect of rotating ${180}^{\circ }$ on the given equation.

Answer to Problem 41E

Here, the equation for the hyperbola is $\frac{y{\text{'}}^2}{a^2}-\frac{x{\text{'}}^2}{b^2}=1$ .

Explanation of Solution

Calculation:

Given equation is :

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Substituting the values of $x=-y\text{'}$ and $y=x\text{'}$ , we have

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

  $\Rightarrow \frac{y{\text{'}}^2}{a^2}-\frac{x{\text{'}}^2}{b^2}=1$

Conclusion:

Hence, the equation for the hyperbola is $\frac{y{\text{'}}^2}{a^2}-\frac{x{\text{'}}^2}{b^2}=1$ .

(c)

To determine

To find: The effect of rotating ${180}^{\circ }$ on the given equation.

Answer to Problem 41E

Here, the equation for the circle $x{\text{'}}^2+y{\text{'}}^2=a^2$ .

Explanation of Solution

Calculation:

given equation is:

  $x^2+y^2=a^2$

Substituting the values of $x=-y\text{'}$ and $y=x\text{'}$ , we have

  $\begin{array}{l}{\left(-y\text{'}\right)}^2+{\left(x\text{'}\right)}^2=a^2\\ \Rightarrow x{\text{'}}^2+y{\text{'}}^2=a^2\end{array}$

Conclusion:

Hence, the equation for the circle $x{\text{'}}^2+y{\text{'}}^2=a^2$ .

(d)

To determine

To find: The effect of rotating ${180}^{\circ }$ on the given equation.

Answer to Problem 41E

Here, the equation for the line $y\text{'}=-\frac{1}{m}x\text{'}$ .

Explanation of Solution

Calculation:

Given equation is :

  $y=mx$

Substituting the values of $x=-y\text{'}$ and $y=x\text{'}$ , we have

  $\left(x\text{'}\right)=m\left(-y\text{'}\right)$

  $\Rightarrow y\text{'}=-\frac{1}{m}x\text{'}$

Conclusion:

Hence, theequation for the line $y\text{'}=-\frac{1}{m}x\text{'}$ .

(e)

To determine

To find: The effect of rotating ${180}^{\circ }$ on the given equation.

Answer to Problem 41E

Here, the equation for the line $y\text{'}=-\frac{1}{m}x\text{'}+\frac{b}{m}$.

Explanation of Solution

Calculation:

(e). Given equation is:

  $y=mx+b$

Substituting the values of $x=-y\text{'}$ and $y=x\text{'}$ , we have

  $\begin{array}{l}\left(x\text{'}\right)=m\left(-y\text{'}\right)+b\\ \Rightarrow y\text{'}=-\frac{1}{m}x\text{'}+\frac{b}{m}\end{array}$

Conclusion:

Therefore, the equation for the line $y\text{'}=-\frac{1}{m}x\text{'}+\frac{b}{m}$.