Chapter 10.7, Problem 27E

To determine

To draw:

The graph of each equation and the value of $\theta$ to the nearest degree.

Answer to Problem 27E

  $\theta ={13.28}^0$ .

Explanation of Solution

Given:

The quadratic equation

  $x^2-xy-4y^2-x-y+4=0$ .

Concept used:

For the general equation

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

The angle rotating $\theta$ about the origin

  $\theta =\frac{\pi }{4}$ If $A=C$

  $\tan 2\theta =\frac{B}{A-C},\text{if}A\ne C$

The number is the square root of the quadratic formula.

  $B^2-4AC$

Calculation:

The quadratic equation

  $x^2-xy-4y^2-x-y+4=\mathrm{0.......................}\left(1\right)$

Comparing the quadratic equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

  $A=1,B=-1,C=-1$

The angle rotating $\theta$ about the origin

  $\theta =\frac{\pi }{4}$ If $A=C$

  $\begin{array}{l}\tan 2\theta =\frac{B}{A-C},\text{if}A\ne C\\ \tan 2\theta =\frac{-1}{1+1}\\ \tan 2\theta =-\frac{1}{2}\\ 2\theta ={\tan }^{-1}\left(\frac{1}{2}\right)\\ 2\theta ={\tan }^{-1}\left(0.5\right)\\ \theta =\frac{{26.56}^0}{2}\\ \theta ={13.28}^0\end{array}$

Draw the graph

  $x^2-xy-4y^2-x-y+4=0$