Chapter 11.5, Problem 20E

a.

To determine

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

Answer to Problem 20E

The graph of the equation is ellipse.

Explanation of Solution

Given information:

The equation is $21x^2+10\sqrt{3}xy+31y^2=144$

Calculation: to find the discriminant,

  $\begin{array}{l}21x^2+10\sqrt{3}xy+31y^2=144\\ A=21,B=10\sqrt{3},C=31\end{array}$

Now, the discriminant is,

  $\begin{array}{l}{B}^2-4AC={\left(10\sqrt{3}\right)}^2-4\left(21\right)\left(31\right)\\ B^2-4AC=300-2604\text{ [multiply]}\\ B^2-4AC<0\end{array}$

Thus, the graph of the equation is ellipse.

b.

To determine

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

Answer to Problem 20E

The equation is $\frac{77}{2}{X}^2+\frac{27}{2}{Y}^2=144$ .

Explanation of Solution

Given information:

The equation is $21x^2+10\sqrt{3}xy+31y^2=144$

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

  $\begin{array}{l}\cos 2\varphi =\frac{A-C}{B}\\ \cos 2\varphi =\frac{21-31}{10\sqrt{3}}\text{ [substitute]}\\ \cos 2\varphi =-\frac{1}{\sqrt{3}}\text{ [simplify]}\end{array}$

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,

  $\begin{array}{l}21{\left(\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\right)}^2+10\sqrt{3}\left(\frac{\sqrt{3}}{2}X-\frac{1}{2}Y\right)\left(\frac{1}{2}X+\frac{\sqrt{3}}{2}Y\right)+31{\left(\frac{1}{2}X+\frac{\sqrt{3}}{2}Y\right)}^2=144\\ 21\left(\frac{3}{4}{X}^2+\frac{1}{4}{Y}^2-\frac{\sqrt{3}}{2}XY\right)+10\sqrt{3}\left(\frac{\sqrt{3}}{4}{X}^2-\frac{\sqrt{3}}{4}{Y}^2+\frac{1}{2}XY\right)+31\left(\frac{1}{4}{X}^2+\frac{3}{4}{Y}^2+\frac{\sqrt{3}}{2}XY\right)=144\text{ [mutliply]}\\ \frac{77}{2}{X}^2+\frac{27}{2}{Y}^2=144\text{ [simplify]}\end{array}$

c.

To determine

To graph: the equation.

Explanation of Solution

Given information:

The equation is $21x^2+10\sqrt{3}xy+31y^2=144$

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

$x$	$y$
$0$	$2.15$
$2.6$	$0$
$-2.6$	$0$

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 ellipse.