Chapter 11.5, Problem 34E

a.

To determine

To show: that the following equation represents a parabola using rotation of axis.

Answer to Problem 34E

  $4.\frac{1}{16}(Y-(-4))={(X-1)}^2$ is the standard equation for an up-down parabola.

Explanation of Solution

Given information: Given equation is $2\sqrt{2}{(x+y)}^2=7x+9y.$

Calculation:

Need to use rotation of axes to show that the equation,

  $2\sqrt{2}{(x+y)}^2=7x+9y.$ , represents a parabola.

  $\begin{array}{l}$ 2\sqrt{2}{(x+y)}^2=7x+9y.$2\sqrt{2}(x^2+y^2+2xy)-7x-9y=0.\\ 2\sqrt{2}{x}^2+4\sqrt{2}xy+2\sqrt{2}{y}^2-7x-9y=0\end{array}$

For that first have to eliminate xy term from the given equation.

On comparing the given equation with: $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\Rightarrow A=2\sqrt{2},B=4\sqrt{2},C=2\sqrt{2}.$

Rotate the axes through the acute angle $\varphi$ then:

  $\begin{array}{c}\cot 2\varphi =\frac{A-C}{B}=\frac{2\sqrt{2}-2\sqrt{2}}{4\sqrt{2}}=\frac{0}{4\sqrt{2}}=0then:\\ \cot 2\varphi =0=\cot {90}^{\circ }\\ 2\varphi ={90}^{\circ }\\ \varphi ={45}^{\circ }\\ \cos {45}^{\circ }=\frac{1}{\sqrt{2}}.\\ \sin {45}^{\circ }=\frac{1}{\sqrt{2}}.\end{array}$

The Rotation of Axes Formula then give:

  $x=\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y\text{and}y=\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y$ .

Substituting into the given equation:

  $\begin{array}{l}2\sqrt{2}{(}^{\frac{1}{\sqrt{2}}}+4\sqrt{2}(\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y)(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y)+2\sqrt{2}{(}^{\frac{1}{\sqrt{2}}}-7(\frac{1}{\sqrt{2}}X-\frac{1}{\sqrt{2}}Y)\\ -9(\frac{1}{\sqrt{2}}X+\frac{1}{\sqrt{2}}Y)=0\\ 4\sqrt{2}{X}^2-8\sqrt{2}X-\sqrt{2}Y=0\\ X^2-2X-\frac{1}{4}Y=0\\ \\ $ 4.\frac{1}{16}(Y-(-4))={(X-1)}^2$\end{array}$

Therefore,

  $4.\frac{1}{16}(Y-(-4))={(X-1)}^2$ is the standard equation for an up-down parabola.

Hence Proved.

b.

To determine

To find: the XY − and xy - coordinates of the vertex and focus.

Answer to Problem 34E

Vertex is (1, -4).

Focus is $(1,-\frac{63}{16})$ .

Explanation of Solution

Given information: Given parabola is $4.\frac{1}{16}(Y-(-4))={(X-1)}^2$ .

Calculation:

  $4.\frac{1}{16}(Y-(-4))={(X-1)}^2$ is the standard equation for an up-down parabola.

Therefore, vertex is ( h, k) = (1,-4) and focal length p = $\frac{1}{16}$

Parabola is symmetric around y- axis and so the focus lies a distance p from the center

(1, -4+ p ) along the y −axis:

  $(1,-4+p)=(1,-4+\frac{1}{16})\Rightarrow (1,-\frac{63}{16})$

c.

To determine

To find: the equation of the directrix in XY −and xy −coordinates.

Answer to Problem 34E

Equation of directrix is:

  $y=-\frac{65}{16}$ .

Explanation of Solution

Given information: Given parabola is $4.\frac{1}{16}(Y-(-4))={(X-1)}^2$

Calculation:

Parabola is symmetric around the y −axis and so the directrix is a parallel to the x -axis, distance p from the center (1, -4) , y − coordinate :

  $\begin{array}{l}y=-4-p\\ y=-4-\frac{1}{16}.\\ y=-\frac{65}{16}.\end{array}$