Chapter 10.7, Problem 22E

To determine

To find:

The equation of the ellipse to given rotation.

Answer to Problem 22E

  ${\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2=-8$ .

Explanation of Solution

Given:

The equation

  $xy=-8,\theta =\frac{\pi }{4}$ .

Concept used:

The general equation of conic section with respect to rotation of $\theta$

  $\begin{array}{l}x\text{'}\cos \theta +y\text{'}\sin \theta \left(x-axis\right)\\ -x\text{'}\sin \theta +y\text{'}\cos \theta \left(y-axis\right)\end{array}$

Calculation:

The equation

  $xy=-8,\theta =\frac{\pi }{4}\mathrm{.......................}\left(1\right)$

Replace $x$ with $x\text{'}$ and $y$ with $-x\text{'}$

  $\begin{array}{l}x\text{'}\cos \theta +y\text{'}\sin \theta \\ -x\text{'}\sin \theta +y\text{'}\cos \theta \\ \theta ={45}^0\\ x\text{'}\cos {45}^0+y\text{'}\sin {45}^0\\ -x\text{'}\sin {45}^0+y\text{'}\cos {45}^0\\ x\text{'}\frac{\sqrt{2}}{2}+y\text{'}\frac{\sqrt{2}}{2}\\ -x\text{'}\frac{\sqrt{2}}{2}+y\text{'}\frac{\sqrt{2}}{2}\end{array}$

Putting these value in equation (1)

  $\begin{array}{l}xy=-8\\ \left(x\text{'}\frac{\sqrt{2}}{2}+y\text{'}\frac{\sqrt{2}}{2}\right)\left(-x\text{'}\frac{\sqrt{2}}{2}+y\text{'}\frac{\sqrt{2}}{2}\right)=-8\\ {\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2+x\text{'}y\text{'}-x\text{'}y\text{'}=-8\\ {\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2=-8\end{array}$

Draw the graph

  $xy=-8$

  ${\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2=-8$