Chapter 8, Problem 36RE

Solution

a)

To identify: the type of conic.

The second degree equation $91x^2-24xy+84y^2-300=0$ represents an ellipse.

Given information:

The second degree equation is given as $91x^2-24xy+84y^2-300=0$ .

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 $A,B,C$ from the second degree equation $91x^2-24xy+84y^2-300=0$ .

  $A=91,B=-24,C=84,$

Calculate $B^2-4AC$ .

  $\begin{array}{c}{B}^2-4AC={\left(-24\right)}^2-4\left(91\right)\left(84\right)\\ =-3000<0\end{array}$

Therefore, the second degree equation $91x^2-24xy+84y^2-300=0$ represents an ellipse.

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 $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

Given information:

The second degree equation is given as $91x^2-24xy+84y^2-300=0$ .

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 $\theta$ .

  $\begin{array}{c}\cot 2\theta =\frac{91-84}{-24}=\frac{-7}{24}\\ \cos 2\theta =\frac{-7}{\sqrt{{\left(-24\right)}^2+7^2}}=-\frac{7}{25}\end{array}$

Use the half-angle identities to compute $\cos \theta$ and $\sin \theta$ .

  $\begin{array}{c}\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}\\ =\sqrt{\frac{1+\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

  $\begin{array}{c}\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}\\ =\sqrt{\frac{1-\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{16}{25}}\\ =\frac{4}{5}\end{array}$

Calculate axis rotation formula.

  $\begin{array}{l}x=ucos\theta -v\sin \theta =\frac{3u}{5}-\frac{4v}{5}\\ y=u\sin \theta +v\cos \theta =\frac{4u}{5}+\frac{3v}{5}\end{array}$

Therefore, the rotational formulas are $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

c)

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

The $\left(u,v\right)$ equation in standard form is $\frac{u^2}{4}+\frac{v^2}{3}=1$ .

Given information:

The second degree equation is given as $91x^2-24xy+84y^2-300=0$ .

Formula used:

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

Calculation:

It is calculated that $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

Substitute the value of $x$ and $y$ in the second degree equation $91x^2-24xy+84y^2-300=0$ .

  $\begin{array}{c}91x^2-24xy+84y^2-300=0\\ 91{\left(\frac{3u}{5}-\frac{4v}{5}\right)}^2-24\left(\frac{3u}{5}-\frac{4v}{5}\right)\left(\frac{4u}{5}+\frac{3v}{5}\right)+84{\left(\frac{4u}{5}+\frac{3v}{5}\right)}^2=300\\ 91\left(\frac{9u^2-24uv+16v^2}{25}\right)-24\left(\frac{12u^2-7uv-12v^2}{25}\right)+84\left(\frac{16u^2+24uv+9v^2}{25}\right)=300\\ 819u^2-2184uv+1456v^2-288u^2+168uv+288v^2+1344u^2+2016uv+756v^2=7500\\ 75u^2+100v^2=300\\ \frac{u^2}{4}+\frac{v^2}{3}=1\end{array}$

Therefore. the $\left(u,v\right)$ equation in standard form is $\frac{u^2}{4}+\frac{v^2}{3}=1$ .

d)

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

The vertices and foci in $\left(u,v\right)$ coordinates are $\left(\pm 2,0\right)$ and $\left(\pm 1,0\right)$ .

Given information:

The second degree equation is given as $91x^2-24xy+84y^2-300=0$ .

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 $\frac{u^2}{4}+\frac{v^2}{3}=1$ .

The vertices of the ellipse are given by $\left(\pm a,0\right)$ and the foci are $\left(\pm c,0\right)$ .

So, the vertices and foci are $\left(\pm 2,0\right)$ and $\left(\pm \sqrt{4-3},0\right)=\left(\pm 1,0\right)$ .

Therefore, the vertices and foci in $\left(u,v\right)$ coordinates are $\left(\pm 2,0\right)$ and $\left(\pm \sqrt{4-3},0\right)=\left(\pm 1,0\right)$ .

e)

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

The vertices of the ellipse in $\left(x,y\right)$ coordinate are $\left(-1.2,-1.6\right)$ and $\left(1.2,1.6\right)$ and the foci of the ellipse in $\left(x,y\right)$ coordinate are $\left(-0.6,-0.8\right)$ and $\left(0.6,0.8\right)$ .

Given information:

The second degree equation is given as $91x^2-24xy+84y^2-300=0$ .

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 $\frac{u^2}{4}+\frac{v^2}{3}=1$ .

The vertices of the ellipse are given by $\left(\pm a,0\right)$ and the foci are $\left(\pm c,0\right)$ .

So, the vertices and foci are $\left(\pm 2,0\right)$ and $\left(\pm \sqrt{4-3},0\right)=\left(\pm 1,0\right)$ .

It is calculated that $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$

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

  $\begin{array}{l}x=\frac{6}{5}=1.2\\ y=\frac{8}{5}=1.6\end{array}$

Substitute $u=-2$ and $v=0$ to calculate vertex.

  $\begin{array}{l}x=-\frac{6}{5}=-1.2\\ y=-\frac{8}{5}=-1.6\end{array}$

So, the vertices of the ellipse in $\left(x,y\right)$ coordinate are $\left(-1.2,-1.6\right)$ and $\left(1.2,1.6\right)$ .

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

  $\begin{array}{l}x=\frac{3}{5}=0.6\\ y=\frac{4}{5}=0.8\end{array}$

Substitute $u=-1$ and $v=0$ to calculate focus.

  $\begin{array}{l}x=-\frac{3}{5}=-0.6\\ y=-\frac{4}{5}=-0.8\end{array}$

So, the foci of the ellipse in $\left(x,y\right)$ coordinate are $\left(-0.6,-0.8\right)$ and $\left(0.6,0.8\right)$ .

Therefore, the vertices of the ellipse in $\left(x,y\right)$ coordinate are $\left(-1.2,-1.6\right)$ and $\left(1.2,1.6\right)$ and the foci of the ellipse in $\left(x,y\right)$ coordinate are $\left(-0.6,-0.8\right)$ and $\left(0.6,0.8\right)$ .