Chapter 8, Problem 35RE

Solution

a)

To identify:the type of conic.

The second degree equation $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ represents a parabola.

Given information:

The second degree equation is given as $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

Formula used:

The second degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ graphs as:

1. A hyperbola if $B^2-4AC>0$ .
2. A parabola if $B^2-4AC=0$ .
3. An ellipse if $B^2-4AC<0$ except for degenerate cases.

Calculation:

Calculate $A,B,C,D,EandF$ from the second degree equation $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

  $A=1,B=2,C=1,D=4\sqrt{2},E=-4\sqrt{2},F=8$

Calculate $B^2-4AC$ .

  $B^2-4AC=2^2-4\times 1\times 1=0$

Therefore, the second degree equation $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ represents a parabola.

b)

To find:the rotation formulas that will convert the $\left(x,y\right)$ equation to a $\left(u,v\right)$ equation with no cross product tern.

The rotational formulas are $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Given information:

The second degree equation is given as $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

Formula used:

For the second degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$:

  $\cot 2\theta =\frac{A-C}{B}where0<\theta <\frac{\pi }{2}$

The axis rotation formulas are given below:

  $\begin{array}{l}x=ucos\theta -v\sin \theta \\ y=u\sin \theta +v\cos \theta \end{array}$

Calculation:

Calculate $\theta$ .

  $\begin{array}{c}\cot 2\theta =\frac{1-1}{2}=0\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

Calculate axis rotation formula.

  $\begin{array}{l}x=ucos\frac{\pi }{4}-v\sin \frac{\pi }{4}=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\\ y=u\sin \frac{\pi }{4}+v\cos \frac{\pi }{4}=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\end{array}$

Therefore, the rotational formulas are $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

c)

To write:the $\left(u,v\right)$ equation in standard form.

The $\left(u,v\right)$ equation in standard form is $u^2=4\left(v-1\right)$

Given information:

The second degree equation is given as $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

Formula used:

Substitute the value of $x$ and $y$ in the second degree equation and further solve the equation.

Calculation:

It is calculated that $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Substitute the value of $x$ and $y$ in the second degree equation $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

  $\begin{array}{c}{\left(\frac{u-v}{\sqrt{2}}\right)}^2+2\left(\frac{u-v}{\sqrt{2}}\right)\left(\frac{u+v}{\sqrt{2}}\right)+{\left(\frac{u+v}{\sqrt{2}}\right)}^2+4\sqrt{2}\left(\frac{u-v}{\sqrt{2}}\right)-4\sqrt{2}\left(\frac{u+v}{\sqrt{2}}\right)+8=0\\ \frac{{\left(u-v\right)}^2}{2}+\left(u^2-v^2\right)+\frac{{\left(u+v\right)}^2}{2}-8v+8=0\\ 4u^2-16v+16=0\\ u^2-4v+4=0\\ u^2=4\left(v-1\right)\end{array}$

Therefore. the $\left(u,v\right)$ equation in standard form is $u^2=4\left(v-1\right)$ .

d)

To find:the vertices and foci in $\left(u,v\right)$ coordinates.

The vertex and focus in $\left(u,v\right)$ coordinates are $\left(0,1\right)$ and $\left(0,2\right)$ .

Given information:

The second degree equation is given as $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

Formula used:

Calculate the vertex and foci in $\left(u,v\right)$ form and then use rotation formulas to compute vertex and foci in $\left(x,y\right)$ coordinate.

Calculation:

The $\left(u,v\right)$ equation in standard form is $u^2=4\left(v-1\right)$ .

From the equation, $h=0andk=1$ .

The vertex of the parabola is $\left(0,1\right)$ and the focus is $\left(0,2\right)$ .

Therefore, the vertex and focus in $\left(u,v\right)$ coordinates are $\left(0,1\right)$ and $\left(0,2\right)$ .

e)

To find:the vertices and foci in $\left(x,y\right)$ coordinates.

The vertex and focus in $\left(x,y\right)$ coordinates are $\left(\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ and $\left(-\sqrt{2},\sqrt{2}\right)$

Given information:

The second degree equation is given as $x^2+2xy+y^2+4\sqrt{2}x-4\sqrt{2}y+8=0$ .

Formula used:

Calculate the vertex and foci in $\left(u,v\right)$ form and then use rotation formulas to compute vertex and foci in $\left(x,y\right)$ coordinate.

Calculation:

The $\left(u,v\right)$ equation in standard form is $u^2=4\left(v-1\right)$ .

From the equation, $h=0andk=1$ .

The vertex of the parabola is $\left(0,1\right)$ and the focus is $\left(0,2\right)$ .

It is calculated that $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Substitute $u=0$ and $v=1$ to calculate vertex.

  $\begin{array}{l}x=\frac{0}{\sqrt{2}}-\frac{1}{\sqrt{2}}=\frac{-1}{\sqrt{2}}=\frac{-\sqrt{2}}{2}\\ y=\frac{0}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}==\frac{\sqrt{2}}{2}\end{array}$

So, the vertex of the parabola in $\left(x,y\right)$ coordinate is $\left(\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ .

Substitute $u=0$ and $v=2$ to calculate focus .

  $\begin{array}{l}x=\frac{0}{\sqrt{2}}-\frac{2}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\\ y=\frac{0}{\sqrt{2}}+\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}==\sqrt{2}\end{array}$

So, the focus of the parabola in $\left(x,y\right)$ coordinate is $\left(-\sqrt{2},\sqrt{2}\right)$ .

Therefore, the vertex and focus in $\left(x,y\right)$ coordinates are $\left(\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ and $\left(-\sqrt{2},\sqrt{2}\right)$ .