Chapter 11.5, Problem 25E

a.

To determine

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

Answer to Problem 25E

The graph of the equation is hyperbola.

Explanation of Solution

Given information:

The equation is $2\sqrt{3}{x}^2-6xy+\sqrt{3}x+3y=0$

Calculation: to find the discriminant,

  $\begin{array}{l}2\sqrt{3}{x}^2-6xy+\sqrt{3}x+3y=0\\ A=2\sqrt{3},B=-6,C=0\end{array}$

Now, the discriminant is,

  $\begin{array}{l}{B}^2-4AC={\left(-6\right)}^2-4\left(2\sqrt{3}\right)\left(0\right)\\ B^2-4AC=36-0\\ B^2-4AC>0\end{array}$

Thus, the graph of the equation is hyoerbola.

b.

To determine

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

Answer to Problem 25E

The equation is ${\left(X-1\right)}^2-3Y^2=1$ .

Explanation of Solution

Given information:

The equation is $2\sqrt{3}{x}^2-6xy+\sqrt{3}x+3y=0$

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

  $\begin{array}{l}\cos 2\varphi =\frac{A-C}{B}\\ \cos 2\varphi =\frac{2\sqrt{3}-0}{-6}\text{ [substitute]}\\ \cos 2\varphi =-\frac{1}{\sqrt{3}}\text{ [simplify]}\end{array}$

Thus, $2\varphi ={60}^{\circ }$ and hence

  $\varphi ={30}^{\circ }$ , so

  $\cos \varphi =\frac{\sqrt{3}}{2}\text{ and sin}\varphi =\frac{1}{2}$

The rotation of axes formulas,

  $\begin{array}{l}x=\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\\ y=\frac{1}{2}X+\frac{\sqrt{3}}{2}Y\end{array}$

Now, substitute in the equation,

  $\begin{array}{l}2\sqrt{3}{\left(\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\right)}^2-6\left(\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\right)\left(\frac{1}{2}X+\frac{\sqrt{3}}{2}Y\right)+\sqrt{3}\left(\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\right)+3\left(\frac{1}{2}X+\frac{\sqrt{3}}{2}Y\right)=0\\ {\left(X-1\right)}^2-3Y^2=1\text{ [by expnading and simplify]}\end{array}$

c.

To determine

To graph: the equation.

Explanation of Solution

Given information:

The equation is $2\sqrt{3}{x}^2-6xy+\sqrt{3}x+3y=0$

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

$x$	$y$
$0$	$0$
$0.8$	$2$
$-0.5$	$0$

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 hyperbola.