Chapter 9, Problem 8CT

a.

To determine

To find: Number of degrees axis must be rotated to eliminate the $xy$ term of the Conic.

Answer to Problem 8CT

  $\theta ={45}^0$

Explanation of Solution

Given information:

  $x^2+6xy+y^2-6=0$

If any conic in the form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Then angle must be rotated to remove $xy$ term is

  $\cot \left(2\theta \right)=\frac{A-C}{B}$

Given

  $x^2+6xy+y^2-6=0$

On comparing with standard equation

We get

  $\begin{array}{l}A=1\\ B=6\\ C=1\end{array}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow \cot \left(2\theta \right)=\frac{1-1}{6}\\ \Rightarrow \cot \left(2\theta \right)=\frac{0}{6}\\ \Rightarrow \cot \left(2\theta \right)=0\\ \Rightarrow 2\theta =\frac{\pi }{2}\\ \Rightarrow \theta =\frac{\pi }{4}={45}^0\end{array}$

Therefore

Number of degrees axis must be rotated to eliminate $xy$ term is $\theta ={45}^0$

b.

To determine

To find:Graph the conic and use a graphing utility to confirm result.

Explanation of Solution

Given information:

  $x^2+6xy+y^2-6=0$

  $x^2+6xy+y^2-6=0$

  $\theta ={45}^0$

Solve $x$ and $y$ using Substitution formulae

  $\begin{array}{l}x=X\cos \left(\theta \right)-Y\sin \left(\theta \right)\\ y=X\sin \left(\theta \right)+Y\cos \left(\theta \right)\end{array}$

Therefore

  $\begin{array}{l}x=X\cos \left(\theta \right)-Y\sin \left(\theta \right)\\ \Rightarrow x=X\cos \left({45}^0\right)-Y\sin \left({45}^0\right)\\ \Rightarrow x=X\left(\frac{1}{\sqrt{2}}\right)-Y\left(\frac{1}{\sqrt{2}}\right)\\ \Rightarrow x=\frac{2}{\sqrt{2}}\left(X-Y\right)\end{array}$

  $\begin{array}{l}y=X\sin \left(\theta \right)+Y\cos \left(\theta \right)\\ \Rightarrow y=X\sin \left({45}^0\right)+Y\cos \left({45}^0\right)\\ \Rightarrow y=X\left(\frac{1}{\sqrt{2}}\right)+Y\left(\frac{1}{\sqrt{2}}\right)\\ \Rightarrow y=\frac{2}{\sqrt{2}}\left(X+Y\right)\end{array}$

  $\begin{array}{l}x=\frac{2}{\sqrt{2}}\left(X-Y\right)\\ \Rightarrow x^2=\frac{1}{2}\left(X^2-2XY+Y^2\right)\end{array}$

  $\begin{array}{l}y=\frac{2}{\sqrt{2}}\left(X+Y\right)\\ \Rightarrow y^2=\frac{1}{2}\left(X^2+2XY+Y^2\right)\end{array}$

  $\begin{array}{l}xy=\left[\frac{2}{\sqrt{2}}\left(X-Y\right)\right]\left[\frac{2}{\sqrt{2}}\left(X+Y\right)\right]\\ \Rightarrow xy=\frac{1}{2}\left[x^2-y^2\right]\end{array}$

Substitute these values in the given equation

  $x^2+6xy+y^2-6=0$

We get

  $\begin{array}{l}{x}^2+6xy+y^2-6=0\\ \Rightarrow \left[\frac{1}{2}\left(X^2-2XY+Y^2\right)\right]+6\left[\frac{1}{2}\left[x^2-y^2\right]\right]+\left[\frac{1}{2}\left(X^2+2XY+Y^2\right)\right]-6=0\\ \Rightarrow \frac{X^2}{2}-XY+\frac{Y^2}{2}+3X^2-3Y^2+\frac{X^2}{2}+XY+\frac{Y^2}{2}=6\\ \Rightarrow 4X^2-2Y^2=6\\ \Rightarrow \frac{4X^2}{6}-\frac{2Y^2}{6}=1\\ \Rightarrow \frac{2X^2}{3}-\frac{Y^2}{3}=1\\ \Rightarrow \frac{X^2}{\left(\frac{3}{2}\right)}-\frac{Y^2}{3}=1\end{array}$

  $\frac{X^2}{\left(\frac{3}{2}\right)}-\frac{Y^2}{3}=1$ -----------(2)

The graph will be Parabola

  

Red colour lines represent the Hyperbola of equation-(1)

Green colour lines represent the Hyperbola of equation-(2)

So, by graphing Utility it was verified.