Chapter 10.7, Problem 31E

To determine

Tofind:the $\theta$ angle to the nearest degree and identify the graph of the function.

Answer to Problem 31E

The angle is $-{30}^0$ .

Explanation of Solution

Given:

  $2x^2+4\sqrt{3}xy+6y^2+\sqrt{3}x-y=0.$

Concept used:

The general equation for conic section is:

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$

Where $A,B,C,D,E\text{ and }F$ are constant.

As by changing the values of some of the constants, the shape of the corresponding conic also changes.

If $B^2-4AC$ , is less than zero, if a conic exists, it will be either a circle or an ellipse.

If $B^2-4AC$ ,equals to zero, if a conic exists, it will be parabola.

If $B^2-4AC$ id greater than zero, if a conic exists, it will be hyperbola.

Calculation:

Consider the equation of the conic:

  $2x^2+4\sqrt{3}xy+6y^2+\sqrt{3}x-y=0.$

The standard form of a conic is:

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$

Rearrange the above equation in this form:

  $2x^2+\left(4\sqrt{3}\right)xy+6y^2+\left(\sqrt{3}\right)x+\left(-1\right)y=0.$

Compare the coefficients:

  $\begin{array}{l}A=2,B=4\sqrt{3},C=6\\ D=\sqrt{3},E=-1\text{ and }F=0.\end{array}$

Now,

  $\begin{array}{l}{B}^2-4AC={\left(4\sqrt{3}\right)}^2-\mathrm{4.2.6}\\ \Rightarrow 48-48=0.\end{array}$

While $B^2-4AC=0$ , from the table of comparisons of the conics,

  $2x^2+4\sqrt{3}xy+6y^2+\sqrt{3}x-y=0.$

Represents a parabola.

For the standard equation of a conic $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$ the angle of rotation $\theta$ is given by:

  $\tan 2\theta =\frac{B}{A-C}$ .

So, the angle of rotation $\theta$ for $2x^2+4\sqrt{3}xy+6y^2+\sqrt{3}x-y=0$ is:

  $\begin{array}{l}\theta =\frac{1}{2}{\tan }^{-1}\left(-\sqrt{3}\right).\\ \Rightarrow \frac{1}{2}\left(-{60}^0\right).\\ \Rightarrow -{30}^0.\end{array}$

Hence, the angle is $-{30}^0$ .