Chapter 10.7, Problem 24E

To determine

To find:

The equation of the ellipse to given rotation.

Answer to Problem 24E

  ${\left(x\text{'}\right)}^2+11{\left(y\text{'}\right)}^2+10\sqrt{3}x\text{'}y\text{'}=64$ .

Explanation of Solution

Given:

The equation

  $16x^2-4y^2=64$ .

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

  $16x^2-4y^2=\mathrm{64.......................}\left(1\right)$

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)

  $\begin{array}{l}16x^2-4y^2=64\\ 16{\left(x\text{'}\frac{1}{2}+y\text{'}\frac{\sqrt{3}}{2}\right)}^2-4{\left(-x\text{'}\frac{\sqrt{3}}{2}+y\text{'}\frac{1}{2}\right)}^2=64\\ \left({\left(x\text{'}\right)}^2\frac{16}{4}+{\left(y\text{'}\right)}^2\frac{48}{4}+x\text{'}y\text{'}\frac{16\sqrt{3}}{2}\right)-4\left({\left(x\text{'}\right)}^2\frac{3}{4}+{\left(y\text{'}\right)}^2\frac{1}{4}-x\text{'}y\text{'}\frac{\sqrt{3}}{2}\right)=64\\ 4{\left(x\text{'}\right)}^2+12{\left(y\text{'}\right)}^2+8\sqrt{3}x\text{'}y\text{'}-3{\left(x\text{'}\right)}^2-{\left(y\text{'}\right)}^2+2\sqrt{3}x\text{'}y\text{'}=64\\ {\left(x\text{'}\right)}^2+11{\left(y\text{'}\right)}^2+10\sqrt{3}x\text{'}y\text{'}=64\end{array}$

Draw the graph

  $16x^2-4y^2=64$

  ${\left(x\text{'}\right)}^2+11{\left(y\text{'}\right)}^2+10\sqrt{3}x\text{'}y\text{'}=64$