helpppp

Question

For each equation, show the

• Factorization of left side of equation
• Equations of asymptotes
• Corner points of rectangle
• x- or y-intercepts

Then graph the hyperbola and include the asymptotes, fundamental rectangle, and intercepts.

Left Side Factored:   Equations of Asymptotes:   Corner Points:   Intercepts:   x2 – y2 = 1 Left Side Factored:  Equations of Asymptotes:    Corner Points:  Intercepts:

Left Side Factored:  Equations of Asymptotes:  Corner Points:  Intercepts:

 

Answer 1

a)

Factorization of left side of equation :

x² – y²  = 1

Substitute formula : a² - b² = (a+b)(a-b)

( y – x )( y + x ) = 1

Equations of asymptotes :

Compare given equation with standard from ( ) and write the coefficients.

Center (0 , 0) , a = 1 , b = 1

Vertices at ( ±a , 0) = (± 1 , 0 ) = (1 , 0) , ( -1 , 0)

To find the asymptotes use formula : y = ±(b/a)x

⇒  y = ±(1/1)x   ⇒  y = ±x.

So asymptotes are   y = x  ;  y = -x.

Corner Points :

To get the corner points of the fundamental rectangle, substitute x coordinate values of vertices in asymptotes.

Substitute x= 1 in the equation y = x  ⇒  y = 1. Hence Corner point is (1,1).

Substitute x= 1 in the equation y = -x  ⇒  y = -1. Hence Corner point is (1,-1).

Substitute x= -1 in the equation y = x  ⇒  y = -1  . Hence Corner point is (-1,-1).

Substitute x= -1 in the equation y = -x  ⇒  y = -(-1) = 1. Hence Corner point is (-1,1).

The corner points of the fundamental rectangle are (1,1) , (1,-1) , (-1,1) , (-1,-1).

Intercepts :

To get the x-intercepts of hyperbola, substitute y = 0  in given hyperbola equation y² – x² = 1.

x² – 0²  = 1    ⇒    x² = 1 ⇒  x = ±1.

So  x-intercepts are (1,0) , (-1,0).

To get the y-intercepts of hyperbole, substitute x =0  in given hyperbola equation y² – x² = 1.

0² – y²  = 1    ⇒    y² = -1.

Square root of the negative values gives no real values.

y-intercepts does not exist.

Given hyperbola equation  x² – y² = 1 has only x-intercepts. So this is a horizontal hyperbola.

Intercepts are (1,0) , (-1,0).

Graph of the given equation x² – y²  = 1 is

Answer 2

b)

Factorization of left side of equation :

y² – (x²/4)= 1

 y² – (x²/2²)= 1

 y² – (x/2)²= 1

Substitute formula : a² - b² = (a+b)(a-b)

[ y – (x/2) ][ y + (x/2) ] = 1

Equations of asymptotes :

Compare given equation with standard from ( ) and write the coefficients.

Center (0 , 0) , a = 1 , b = 2

Vertices at (0 , ±b) = (0 , ± 2) = (0 , 2) , (0 , -2)

To find the asymptotes use formula : y = ±(a/b)x

⇒  y = ±(1/2)x   ⇒  y = ±x/2.

So asymptotes are   y = x/2  ;  y = -x/2.

Corner Points :

To get the corner points of the fundamental rectangle, substitute y coordinate values of vertices in asymptotes.

Substitute x= 2 in the equation y = x/2  ⇒  y = 2/2=1. Hence Corner point is (2,1).

Substitute x= 2 in the equation y = -x/2  ⇒  y = -2/2=-1. Hence Corner point is (2,-1).

Substitute x= -2 in the equation y = x/2  ⇒  y = -2/2=-1  . Hence Corner point is (-2,-1).

Substitute x= -2 in the equation y = -x/2  ⇒  y = -(-2/2) = 1. Hence Corner point is (-2,1).

The corner points of the fundamental rectangle are (2,1) , (2,-1) , (-2,1) , (-2,-1).

Intercepts :

To get the x-intercepts of hyperbola, substitute y = 0  in given hyperbola equation  y² – (x²/4)= 1.

0² – (x²/4)= 1    ⇒    x² = - 4

Square root of the negative values gives no real values.x-intercepts does not exist.

 

To get the y-intercepts of hyperbole, substitute x =0  in given hyperbola equation  y² – (x²/4)= 1.

 y² – (0²/4)= 1     ⇒    y² = 1 ⇒    y = ±1.

So  y-intercepts are (0,1) , (0,-1).

Given hyperbola equation y² – x² = 1 has only y-intercepts. So this is a vertical hyperbola.

Intercepts are (0,1) , (0,-1).

Graph of the given equation y² – (x²/4)= 1 is

Answer 3

c)

Factorization of left side of equation :

x² – (y²/9)  = 1

x² – (y²/3²)= 1

x² – (y/3)²= 1

Substitute formula : a² - b² = (a+b)(a-b)

[ x – (y/3) ][ x + (y/3) ] = 1

Equations of asymptotes :

Compare given equation with standard from ( ) and write the coefficients.

Center (0 , 0) , a = 1 , b = 3

Vertices at ( ±a , 0) = (± 1 , 0 ) = (1 , 0) , ( -1 , 0)

To find the asymptotes use formula : y = ±(b/a)x

⇒  y = ±(3/1)x   ⇒  y = ±3x.

So asymptotes are   y = 3x  ;  y = -3x.

Corner Points :

To get the corner points of the fundamental rectangle, substitute x coordinate values of vertices in asymptotes.

Substitute x= 1 in the equation y = 3x  ⇒  y = 3. Hence Corner point is (1,3).

Substitute x= 1 in the equation y = -3x  ⇒  y = -3. Hence Corner point is (1,-3).

Substitute x= -1 in the equation y = 3x  ⇒  y = -3  . Hence Corner point is (-1,-3).

Substitute x= -1 in the equation y = -3x  ⇒  y = -(-3) = 3. Hence Corner point is (-1,3).

The corner points of the fundamental rectangle are (1,3) , (1,-3) , (-1,3) , (-1,-3).

Intercepts :

To get the x-intercepts of hyperbola, substitute y = 0  in given hyperbola equation x² – (y²/9)  = 1.

x² – (0²/9)  = 1    ⇒    x² = 1 ⇒  x = ±1.

So  x-intercepts are (1,0) , (-1,0).

To get the y-intercepts of hyperbole, substitute x =0  in given hyperbola equation x² – (y²/9)  = 1.

0² – (y²/9)  = 1    ⇒    y² = -1.

Square root of the negative values gives no real values.

y-intercepts does not exist.

Given hyperbola equation  x² – (y²/9)  = 1 has only x-intercepts. So this is a horizontal hyperbola.

Intercepts are (1,0) , (-1,0).

Graph of the given equation x² – (y²/9)  = 1 is