Chapter 8.4, Problem 38E

Solution

a)

To find: The type of the conic section for the equation $x^2+xy+y^2=3$ .

The graph of equation $x^2+xy+y^2=3$ is an ellipse.

Given information:

The equation is $x^2+xy+y^2=3$ .

Calculation:

Draw the graph of equation $x^2+xy+y^2=3$

  

Therefore, the graph represents an ellipse.

b)

To find: The equation in terms of $\left(u,v\right)$ with no cross-product term and using axis of rotation.

The required equation in terms of $\left(u,v\right)$ is $\frac{v^2}{6}+\frac{u^2}{2}=1$ .

Given information:

The equation is $x^2+xy+y^2=3$ .

Formula used:

Formula used to find the rotation angle $\text{θ}$ is:

  $\cot \theta =\frac{A-C}{B}$ , $0<\theta <\frac{\pi }{2}$

Here, A , B , and C are the values obtain by comparing given equation with equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

The equations for the axis of rotation are,

  $x=u\cos \theta -v\sin \theta$

  $y=u\sin \theta +v\cos \theta$

Calculation:

Comparing the given equation with the general equation,

  $A=1$

  $B=1$

  $C=1$

Calculate the angle of rotation.

  $\begin{array}{c}\cot 2\theta =\frac{1-1}{1}\\ \cot 2\theta =0\\ 2\theta ={\cot }^{-1}\left(0\right)\\ 2\theta =\frac{\pi }{2}\end{array}$

Divide the obtained equation by 2 to obtain the value of $\theta$ .

  $\begin{array}{c}\frac{2\theta }{2}=\frac{\frac{\pi }{2}}{2}\\ \theta =\frac{\pi }{4}\end{array}$

Substitute $\frac{\pi }{4}$ for $\theta$ into the above formulas to the values the values $x$ and $y$ in $\left(u,v\right)$ -coordinates.

  $\begin{array}{c}x=u\cos \left(\frac{\pi }{4}\right)-v\sin \left(\frac{\pi }{4}\right)\\ =u\cdot \frac{1}{\sqrt{2}}-v\cdot \frac{1}{\sqrt{2}}\\ =\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\end{array}$

  $\begin{array}{c}y=u\sin \left(\frac{\pi }{4}\right)+v\cos \left(\frac{\pi }{4}\right)\\ =u\cdot \frac{1}{\sqrt{2}}+v\cdot \frac{1}{\sqrt{2}}\\ =\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\end{array}$

Substituting the values of x and y in given equation and simplify.

  $\begin{array}{c}2{\left(\frac{1}{\sqrt{2}}u-\frac{v}{\sqrt{2}}\right)}^2+\sqrt{3}\left(\frac{1}{\sqrt{2}}u-\frac{v}{\sqrt{2}}\right)\left(\frac{u}{\sqrt{2}}+\frac{1}{\sqrt{2}}v\right)+{\left(\frac{u}{\sqrt{2}}+\frac{1}{\sqrt{2}}v\right)}^2=3\\ \frac{3}{2}{u}^2+\frac{1}{2}{v}^2=3\\ \frac{v^2}{6}+\frac{u^2}{2}=1\end{array}$

The equation $x^2+xy+y^2=3$ transfers in $\left(u,v\right)$ coordinates with no cross product as $\frac{v^2}{6}+\frac{u^2}{2}=1$ .

c)

To find: The vertex or vertices of equation $x^2+xy+y^2=3$ in $\left(u,v\right)$ system.

The vertices of equation $x^2+xy+y^2=3$ in $\left(u,v\right)$ system are $\left(\text{0,}\sqrt{6}\right)$ and $\left(\text{0,-}\sqrt{6}\right)$ .

Given information:

The equation is $x^2+xy+y^2=3$ .

The equation in $\left(u,v\right)$ system is,

  $\frac{v^2}{6}+\frac{u^2}{2}=1$

Formula used:

The standard formula for an ellipse is, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Vertices of ellipse are: $\left(0,\pm b\right)$

Calculation:

Comparing this equation with standard equation of ellipse.

  $\begin{array}{c}{b}^2=6\\ b=\pm \sqrt{6}\end{array}$

Therefore, the vertices of ellipse are $\left(\text{0,}\sqrt{6}\right)$ and $\left(\text{0,-}\sqrt{6}\right)$ ..

d)

To find: The vertices of ellipse in $\left(x,y\right)$ system.

The vertices of equation $x^2+xy+y^2=3$ are $\left(-\sqrt{3},\sqrt{3}\right)$ and $\left(\sqrt{3},-\sqrt{3}\right)$ .

Given information:

The equation is $x^2+xy+y^2=3$ .

The equations for the rotation of axis are,

  $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$

  $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$

The vertices of equation $x^2+xy+y^2=3$ in $\left(u,v\right)$ system are $\left(\text{0,}\sqrt{6}\right)$ and $\left(\text{0,-}\sqrt{6}\right)$ .

Calculation:

Write the equation of the rotation of axis in terms of x and y .

  $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$

  $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$

Substitute the values $u=0$ and $v=\sqrt{6}$ in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{0}{\sqrt{2}}-\frac{\sqrt{6}}{\sqrt{2}}\\ =-\sqrt{3}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{0}{\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}}\\ =\sqrt{3}\end{array}$

Substitute the values $u=0$ and $v=-\sqrt{6}$ in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{0}{\sqrt{2}}-\frac{-\sqrt{6}}{\sqrt{2}}\\ =\sqrt{3}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{0}{\sqrt{2}}+\frac{-\sqrt{6}}{\sqrt{2}}\\ =-\sqrt{3}\end{array}$

The vertices of equation $x^2+xy+y^2=3$ are $\left(-\sqrt{3},\sqrt{3}\right)$ and $\left(\sqrt{3},-\sqrt{3}\right)$ .