Chapter 8.4, Problem 36E

Solution

a)

To find: The type of the conic section for the equation $3xy+15=0$ .

The graph of equation $3xy+15=0$ is a hyperbola.

Given information:

The equation is $3xy+15=0$ .

Calculation:

Draw the graph of equation $3xy+15=0$ .

  

Therefore, the graph represents a hyperbola.

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 $\frac{v^2}{10}-\frac{u^2}{10}=1$ .

Given information:

The equation is $3xy+15=0$ .

Formula used:

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

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

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,

  $A=0$

  $B=3$

  $C=0$

Calculate the angle of rotation.

  $\begin{array}{c}\cot 2\theta =\frac{0-0}{3}\\ \cot 2\theta =0\\ 2\theta ={\cot }^{-1}\left(0\right)\\ 2\theta =\frac{\pi }{2}\end{array}$

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.

  $\begin{array}{c}3\left(\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\right)\left(\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\right)+15=0\\ 3\left(\frac{u^2}{2}-\frac{v^2}{2}\right)=-15\\ \frac{u^2}{2}-\frac{v^2}{2}=-5\\ \frac{v^2}{10}-\frac{u^2}{10}=1\end{array}$

The equation $3xy+15=0$ transfers in $\left(u,v\right)$ coordinates with no cross product as $\frac{v^2}{10}-\frac{u^2}{10}=1$ .

c)

To find: The vertex or vertices of equation $3xy+15=0$ in $\left(u,v\right)$ system.

The vertices of equation $3xy+15=0$ in $\left(u,v\right)$ system are $\left(\text{0,}\sqrt{10}\right)$ and $\left(0,-\sqrt{10}\right)$ .

Given information:

The equation is $3xy+15=0$ .

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

  $\frac{v^2}{10}-\frac{u^2}{10}=1$

Formula used:

The standard formula for a hyperbola is, $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

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

Calculation:

Comparing this equation with standard equation of hyperbola.

  $\begin{array}{c}{b}^2=10\\ b=\pm \sqrt{10}\end{array}$

Therefore, the vertices of hyperbola are $\left(\text{0,}\sqrt{10}\right)$ and $\left(0,-\sqrt{10}\right)$ .

d)

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

The vertices of equation $3xy+15=0$ are $\left(2\sqrt{2},2\sqrt{2}\right)$ and $\left(-2\sqrt{2},-2\sqrt{2}\right)$.

Given information:

The equation is $3xy+15=0$ .

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 $3xy+15=0$ in $\left(u,v\right)$ system are $\left(\text{0,}\sqrt{10}\right)$ and $\left(0,-\sqrt{10}\right)$ .

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 $v=\sqrt{10}$ in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{0}{\sqrt{2}}-\frac{\sqrt{10}}{\sqrt{2}}\\ =-\sqrt{5}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{0}{\sqrt{2}}+\frac{\sqrt{10}}{\sqrt{2}}\\ =\sqrt{5}\end{array}$

Substitute the values $u=0$ and $v=\sqrt{10}$ in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{0}{\sqrt{2}}-\frac{-\sqrt{10}}{\sqrt{2}}\\ =\sqrt{5}\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{0}{\sqrt{2}}+\frac{-\sqrt{10}}{\sqrt{2}}\\ =-\sqrt{5}\end{array}$

The vertices of equation $3xy+15=0$ are $\left(-\sqrt{5},\sqrt{5}\right)$ and $\left(\sqrt{5},-\sqrt{5}\right)$ .