helppp plz

Question

1. The graph of y2 – x2 = 1 is a _______________ .
2. It can be factored as (y – ___ )( ___ + x) = 1.
3. The graphs of (y – x) = 0 and (y + x) = 0 are both _________ .
   
   

   line 1: 

   line 2: 

   These lines are the asymptotes of the hyperbola.

   1. Write the equations of these two lines.
   2. Find the corner points of the fundamental rectangle.
4. Sketch the two lines and the fundamental rectangle on the grid.
5. x-intercepts:

   y-intercepts:

   Therefore, this is a vertical hyperbola and the points where the fundamental rectangle intersects the y-axis, (____, ____) and (____, ____) , are the vertices of the hyperbola.

   Use this information to sketch the hyperbola .Find the intercepts:

Answer 1

3)

Given equation : y² – x² = 1

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

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

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

 

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

⇒  y = ±(1/1)x   ⇒  y = ±x.

So asymptotes are   y = x  ;  y = -x

⇒  (y – x) = 0  ;  (y + x) = 0

The graphs of (y – x) = 0 and (y + x) = 0 are asymptotes.

Answer 2

3a)

The equations of two asymptote lines are

Line 1 : (y – x) = 0

Line 2 : (y + x) = 0

Answer 3

3b)

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

Substitute y= 1 in the equation (y – x) = 0  ⇒  1-x = 0  ⇒ x = 1.Hence Corner point is (1,1).

Substitute y= 1 in the equation (y + x) = 0  ⇒  1+x = 0  ⇒ x = -1. Hence Corner point is (-1,1).

Substitute y= -1 in the equation (y – x) = 0  ⇒  -1-x = 0  ⇒ x = -1. Hence Corner point is (-1,-1).

Substitute y= -1 in the equation (y + x) = 0  ⇒  -1+x = 0  ⇒ x = 1. Hence Corner point is (1,-1).

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

Answer 4

4)

Graph of two asymptote lines ( y=x , y=-x ) and the fundamental rectangle with corner points (1,1) , (-1,1) , (-1,-1) , (1,-1) is

Answer 5

5)

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

0² – x² = 1    ⇒    x² = –1.

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 = 1.

y² – 0² = 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.

Solution : Given equation is vertical hyperbola equation and the points where the fundamental rectangle intersects the y-axis, (0,1) , (0,-1) are the vertices of the hyperbola.

Answer 6

1)

Equation y² – x² = 1 is in the standard form of vertical hyperbola with (0,0) center.

The graph of y² – x² = 1 is a Hyperbola.

 

Answer 7

2)

y² – x² =1

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

(y – x )( y + x) = 1

It can be factored as (y – x )( y + x) = 1.