Chapter 10.7, Problem 23E

To determine

To find:

The equation of the ellipse to given rotation.

Answer to Problem 23E

  ${\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2-x\text{'}\frac{5}{2}-y\text{'}\frac{5\sqrt{3}}{2}=3$ .

Explanation of Solution

Given:

The ellipse equation

  $x^2-5x+y^2=3$ .

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 ellipse equation

  $x^2-5x+y^2=\mathrm{3.......................}\left(1\right)$

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

  $\begin{array}{l}x\text{'}\cos \theta +y\text{'}\sin \theta \\ -x\text{'}\sin \theta +y\text{'}\cos \theta \\ \theta ={60}^0\\ x\text{'}\cos {60}^0+y\text{'}\sin {60}^0\\ -x\text{'}\sin {60}^0+y\text{'}\cos {60}^0\\ x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\\ -x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\end{array}$

Putting these value in equation (1)

  $\begin{array}{l}{x}^2-5x+y^2=3\\ {\left(x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\right)}^2+{\left(-x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\right)}^2-5\left(x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\right)=3\\ \left({\left(x\text{'}\right)}^2\frac{1}{4}+{\left(y\text{'}\right)}^2\frac{3}{4}+x\text{'}y\text{'}\frac{\sqrt{3}}{2}\right)+\left({\left(x\text{'}\right)}^2\frac{3}{4}+{\left(y\text{'}\right)}^2\frac{1}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}\right)-5\left(x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\right)=3\\ {\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2-x\text{'}\frac{5}{2}-y\text{'}\frac{5\sqrt{3}}{2}=3\end{array}$

Draw the graph

  $x^2-5x+y^2=3$

  ${\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2-x\text{'}\frac{5}{2}-y\text{'}\frac{5\sqrt{3}}{2}=3$