Chapter 11.5, Problem 33E

a.

To determine

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

Answer to Problem 33E

  $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ is the standard equation for a right-left facing hyperbola.

Explanation of Solution

Given information: Given equation is $7x^2+48xy-7y^2-200x-150y+600=0.$

Calculation:

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

  $7x^2+48xy-7y^2-200x-150y+600=0$ , represents a hyperbola.

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

On comparing the given equation with: $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\Rightarrow A=7,B=48,C=-7.$

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

  $\begin{array}{c}\cot 2\varphi =\frac{A-C}{B}=\frac{7-(-7)}{48}=\frac{14}{48}=\frac{7}{24}then:\\ Base=7.\\ Perpendicular=24.\\ Hypontenuse=\sqrt{7^2+{24}^2}=25.\\ \cos 2\varphi =\frac{7}{25}\\ \text{Using}\text{the}\text{half}\text{angle}\text{formula:}\\ \cos \varphi =\sqrt{\frac{1+\frac{7}{25}}{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}.\\ \sin \varphi =\sqrt{\frac{1-\frac{7}{25}}{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\end{array}$

The Rotation of Axes Formula then give:

  $x=\frac{4}{5}X-\frac{3}{5}Y\text{and}y=\frac{3}{5}X+\frac{4}{5}Y$ .

Substituting into the given equation:

  $\begin{array}{l}$ 7{(\frac{4}{5}X-\frac{3}{5}Y)}^2+48(\frac{4}{5}X-\frac{3}{5}Y)(\frac{3}{5}X+\frac{4}{5}Y)-7{(\frac{3}{5}X+\frac{4}{5}Y)}^2-200(\frac{4}{5}X-\frac{3}{5}Y)-150(\frac{3}{5}X+\frac{4}{5}Y)+600=0$\frac{625}{25}{X}^2-\frac{625}{25}{Y}^2-\frac{1250}{5}X+600=0\\ 25X^2-25Y^2-250X+600=0\\ X^2-Y^2-10X+24=0\\ X^2-10X+25-Y^2-1=0\\ X^2-10X+25-Y^2=1.\\ \frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1.\end{array}$

Therefore,

  $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ is the standard equation for a right-left facing hyperbola.

Hence Proved.

b.

To determine

To find: the XY − and xy - coordinates of the center, vertices and foci.

Answer to Problem 33E

Center is (5, 0).

Vertices are (6, 0), (4, 0).

Foci are $(5+\sqrt{2},0),(5-\sqrt{2},0)$ .

Explanation of Solution

Given information: Given hyperbola is $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ .

Calculation:

  $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ is the standard equation for a right-left facing hyperbola with center ( h , k ) as (5,0) and semi axis a =1 and semi −conjugate −axis b =1.

The vertices ( h+a ,k ) ,( h-a ,k ) are the two bending points of the hyperbola with center (h ,k) and semi axis a,b.

Given parabola is right-left hyperbola with center ( h , k ) as (5,0) and semi axis a =1 and semi −conjugate −axis b =1.

So, the vertices are (5+1,0) ,(5-1,0) = (6,0) ,(4,0).

Therefore vertices are (6, 0) , (4,0).

For a right −left facing hyperbola, the foci are defined as ( h+c ,k ) ,( h-c ,k ), where $c=\sqrt{a^2+b^2}\Rightarrow \sqrt{1^2+1^2}=\sqrt{2}$ .

So Foci are $(5+\sqrt{2},0),(5-\sqrt{2},0)$ .

c.

To determine

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

Answer to Problem 33E

Equations of asymptotes are:

  $y=x-5,y=-x+5.$

Explanation of Solution

Given information: Given hyperbola is $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ .

Formula used:

For right-left hyperbola the asymptotes are $y=\pm \frac{b}{a}(x-h)+k$ .

Calculation:

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ .

For right-left hyperbola the asymptotes are $y=\pm \frac{b}{a}(x-h)+k$ .

  $\frac{{(X-5)}^2}{1^2}-\frac{{(Y-0)}^2}{1^2}=1$ is the standard equation for a right-left facing hyperbola with center ( h , k) as (5,0) and semi axis a =1 and semi −conjugate −axis b =1.

So, equations of asymptotes are:

  $\begin{array}{l}y=\frac{1}{1}(x-5)+0,y=-\frac{1}{1}(x-5)+0.\\ y=x-5,y=-x+5.\end{array}$