Chapter 10.7, Problem 2CFU

To determine

To find:

The equation of the ellipse to given rotation.

Answer to Problem 2CFU

  ${\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}+x\text{'}\frac{8\sqrt{3}}{2}+y\text{'}\frac{8}{2}=0$ .

Explanation of Solution

Given:

The ellipse equation

  $y^2+8x=0$ .

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:

Counter example:-

The ellipse equation

  $y^2+8x=\mathrm{0.......................}\left(1\right)$

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

  $\begin{array}{l}x\text{'}\cos \theta +y\text{'}\sin \theta \\ -x\text{'}\sin \theta +y\text{'}\cos \theta \\ \theta ={30}^0\\ x\text{'}\cos {30}^0+y\text{'}\sin {30}^0\\ -x\text{'}\sin {30}^0+y\text{'}\cos {30}^0\\ x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\\ -x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\end{array}$

Putting these value in equation (1)

  $\begin{array}{l}8\left(x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\right)+{\left(-x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\right)}^2=0\\ 8\left(x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\right)+{\left({\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}\right)}^2=0\\ \left(x\text{'}\frac{8\sqrt{3}}{2}+y\text{'}\frac{8}{2}\right)+\left({\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}\right)=0\\ {\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}+x\text{'}\frac{8\sqrt{3}}{2}+y\text{'}\frac{8}{2}=0\end{array}$

Draw the graph

  $y^2+8x=0$

  ${\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}+x\text{'}\frac{8\sqrt{3}}{2}+y\text{'}\frac{8}{2}=0$