Chapter 11.5, Problem 18E

(a)

To determine

To check: The graph of the equation $13x^2+6\sqrt{3}xy+7y^2=16$ is a parabola, an ellipse, or a hyperbola.

Answer to Problem 18E

The graph of the equation $13x^2+6\sqrt{3}xy+7y^2=16$ is an ellipse.

Explanation of Solution

Given information:

The given equation is $13x^2+6\sqrt{3}xy+7y^2=16$ .

Calculation:

Write the standard equation for the graph when the graph is degenerated.

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Compare the given equation with the standard equation of the graph.

  $A=13$ , $B=6\sqrt{3}$ , $C=7$

Calculate the discriminant.

  $\begin{array}{c}{B}^2-4AC={\left(6\sqrt{3}\right)}^2-4\times 13\times 7\\ =108-364\\ =-256\\ <0\end{array}$

Since, the discriminant is less than 0. So, the graph of the equation will be an ellipse.

Therefore, the graph of the equation $13x^2+6\sqrt{3}xy+7y^2=16$ is an ellipse.

(b)

To determine

To find: The equation by eliminating the $xy$ -term by use of rotation of axes.

Answer to Problem 18E

The equation is $\left[\begin{array}{l}13{\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X-\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)}^2+6\sqrt{3}\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X-\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X+\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)\\ +7{\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X+\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)}^2=16\end{array}\right]$ .

Explanation of Solution

Given information:

The given equation is $13x^2+6\sqrt{3}xy+7y^2=16$ .

Calculation:

Write the standard equation for the graph when the graph is degenerated.

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Compare the given equation with the standard equation of the graph.

  $A=13$ , $B=6\sqrt{3}$ , $C=7$

Write the equation for an angle through which axes are rotated.

  $\cos 2\varphi =\frac{A-C}{B}$

Substitute the values in the above equation.

  $\begin{array}{c}\cos 2\varphi =\frac{13-7}{6\sqrt{3}}\\ \cos 2\varphi =\frac{6}{6\sqrt{3}}\\ \cos 2\varphi =\frac{1}{\sqrt{3}}\end{array}$

Calculate the value of $\cos \varphi$ and $\sin \varphi$ using half-angle formula.

  $\begin{array}{c}\cos \varphi =\sqrt{\frac{1+\frac{1}{\sqrt{3}}}{2}}\\ =\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}\end{array}$

and,

  $\begin{array}{c}\sin \varphi =\sqrt{\frac{1-\frac{1}{\sqrt{3}}}{2}}\\ =\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}\end{array}$

Write the equation for the rotation of axes.

  $x=\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X-\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y$

and,

  $y=\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X+\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y$

Substitute the value of $x$ and

  $y$ in the equation.

  $\left[\begin{array}{l}13{\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X-\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)}^2+6\sqrt{3}\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X-\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X+\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)\\ +7{\left(\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}X+\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}Y\right)}^2=16\end{array}\right]$

Further, simplify.

  $\left[\begin{array}{l}13\left(\frac{\sqrt{3}+1}{2\sqrt{3}}{X}^2+\frac{\sqrt{3}-1}{2\sqrt{3}}{Y}^2-2\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}XY\right)+6\sqrt{3}\left(\frac{\sqrt{3}+1}{2\sqrt{3}}{X}^2-\frac{\sqrt{3}-1}{2\sqrt{3}}{Y}^2\right)+\\ 7\left(\frac{\sqrt{3}+1}{2\sqrt{3}}{X}^2+\frac{\sqrt{3}-1}{2\sqrt{3}}{Y}^2+2\sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}}\sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}XY\right)=16\end{array}\right]$

On expanding and simplifying.

  $4X^2+Y^2=4$

Therefore, the equation is $4X^2+Y^2=4$ .

(c)

To determine

To plot: The graph of the equation $4X^2+Y^2=4$ .

Explanation of Solution

Given information:

The given equation is $4X^2+Y^2=4$ .

Graph:

The graph of the equation is given below.

  

Interpretation:

The ellipse has center at the origin and vertex at $\left(0,2\right)$