Chapter 11.5, Problem 23E

a.

To determine

To state: that whether the graph of the equation is a parabola, an ellipse or a hyperbola.

Answer to Problem 23E

The graph of the equation is parabola.

Explanation of Solution

Given information:

The equation is $x^2+2xy+y^2+x-y=0$

Calculation: to find the discriminant,

  $\begin{array}{l}{x}^2+2xy+y^2+x-y=0\\ A=1,B=2,C=1\end{array}$

Now, the discriminant is,

  $\begin{array}{l}{B}^2-4AC={\left(2\right)}^2-4\left(1\right)\left(1\right)\\ B^2-4AC=1-1\text{ [multiply]}\\ B^2-4AC=0\end{array}$

Thus, the graph of the equation is parabola.

b.

To determine

To eliminate: the $xy$ term by using rotation of axes.

Answer to Problem 23E

The equation is $Y=\sqrt{2}{X}^2$ .

Explanation of Solution

Given information:

The equation is $x^2+2xy+y^2+x-y=0$

Calculation: to eliminate the $xy$ -term, rotate the axes through an angle $\varphi$ that satisfies,

  $\begin{array}{l}\cos 2\varphi =\frac{A-C}{B}\\ \cos 2\varphi =\frac{1-1}{2}\text{ [substitute]}\\ \cos 2\varphi =0\text{ [simplify]}\end{array}$

Thus, $2\varphi =\frac{\pi }{2}$ so $\varphi =\frac{\pi }{4}$ ,

  $\begin{array}{l}\cos \varphi =\frac{1}{\sqrt{2}}\\ \sin \varphi =\frac{1}{\sqrt{2}}\end{array}$

The rotation of axes formulas,

  $\begin{array}{l}x=\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y\\ y=\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y\end{array}$

Now, substitute in the equation,

  $\begin{array}{l}{\left(\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y\right)}^2+2\left(\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y\right)\left(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y\right)+{\left(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y\right)}^2+\left(\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y\right)-\left(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y\right)=0\\ Y=\sqrt{2}{X}^2\text{ [by expanding and simplify]}\end{array}$

c.

To determine

To graph: the equation.

Explanation of Solution

Given information:

The equation is $x^2+2xy+y^2+x-y=0$

Graph: to draw the graph of the equation, use the plotting point method,

$x$	$y$
$0$	$1$
$0$	$0$
$-1$	$0$

Now, by using the above table and join the points,

The graph can be obtained as:

  

Interpretation: form the above graph it can be observed that the graph of equation is parabola.