Chapter 10.7, Problem 32E

To determine

Tofind:the $\theta$ angle to the nearest degree and identify the conic from the equation.

Answer to Problem 32E

Theangle is ${45}^0$ .

Explanation of Solution

Given:

  $2x^2+4xy+2y^2+2\sqrt{2}x-2\sqrt{2}y+12=0.$

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:

  $2x^2+4xy+2y^2+2\sqrt{2}x-2\sqrt{2}y+12=0.$

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:

  $2x^2+4xy+2y^2+\left(2\sqrt{2}\right)x+\left(-2\sqrt{2}\right)y+12=0.$

Compare the coefficients:

  $\begin{array}{l}A=2,B=4,C=2\\ D=2\sqrt{2},E=-2\sqrt{2}\text{ and }F=12.\end{array}$

Now,

  $\begin{array}{l}{B}^2-4AC={\left(4\right)}^2-\mathrm{4.2.2}\\ \Rightarrow 16-16=0.\end{array}$

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

  $2x^2+4xy+2y^2+2\sqrt{2}x-2\sqrt{2}y+12=0.$

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 $2x^2+4xy+2y^2+2\sqrt{2}x-2\sqrt{2}y+12=0.$ is:

  $\begin{array}{l}\theta =\frac{1}{2}{\tan }^{-1}\left(\frac{4}{0}\right).\\ \Rightarrow \frac{1}{2}\left({90}^0\right).\\ \Rightarrow {45}^0.\end{array}$

Hence, the angle is ${45}^0$ .