Hyperbola: Asymptotes

Before discussing asymptotes of a hyperbola recall that a hyperbola can have a horizontal or a vertical transverse axis.

Let's quickly review the standard form of the hyperbola.

STANDARD EQUATION OF A HYPERBOLA:


➢ Center coordinates (h, k)

➢ a = distance from vertices to the center

➢ c = distance from foci to center

c 2 = a 2 + b 2 b= c 2 a 2


( xh ) 2 a 2 ( yk ) 2 b 2 =1 transverse axis is horizontal


( yk ) 2 a 2 ( xh ) 2 b 2 =1 transverse axis is vertical




A hyperbola has two asymptotes as shown in Figure 1:



The asymptotes pass through the center of the hyperbola (h, k) and intersect the vertices of a rectangle with side lengths of 2a and 2b. The line segment of length 2b joining points (h,k + b) and (h,k - b) is called the conjugate axis.

The equations of the asymptotes are:

EQUATION OF THE ASYMPTOTES OF A HYPERBOLA:


➢ Center coordinates (h, k)

➢ a = distance from vertices to the center

➢ c = distance from foci to center

c 2 = a 2 + b 2 b= c 2 a 2


y=k± b a ( xh ) transverse axis is horizontal


y=k± a b ( xh ) transverse axis is vertical




Let's use these equations in some examples:

Example 1: Find the equations of the asymptotes of the hyperbola
3x2 - 2y2 + 18x + 15 = 0.

Step 1: Group the x- and y-terms on the left-hand side of the equation.

3x2 - 2y2 + 18x + 15 = 0


(3 x 2 +18x)+(2 y 2 )+15=0

Step 2: Move the constant term to the right-hand side.

(3 x 2 +18x)+(2 y 2 )=15

Step 3: Complete the square for the x- and y-groups.

(3 x 2 +18x)+(2 y 2 )=15


Complete the square for the x-group


(3x2 + 18x)


Factor out a 3 so the x2-coefficient is 1


3(x2 + 6x)


Take the coefficient of the x-term, divide by 2 and square the result.


6 2 =3; 3 2 =9


Add the result to the x-group.


3(x2 + 6x + 9)


Complete the square for the y-group


-2y2


Because there is no y-term no work is necessary.


Final result.


3( x 2 +6x+9 )2 y 2 =15+?

Step 4: Add the values added to the left-hand side to the right-hand side.

3( x 2 +6x+9 )2 y 2 =15+3( 9 )


3( x 2 +6x+9 )2 y 2 =12

Step 5: Write the x-group and y-group as perfect squares.

3 ( x+3 ) 2 2 y 2 =12

Step 6: Divide both sides by the value on the right-hand side, simplify and write the denominators as perfect squares.

3 ( x+3 ) 2 12 2 y 2 12 =1


Divide by 12:


( x+3 ) 2 4 y 2 6 =1


Write denominators as perfect squares:


( x+3 ) 2 2 2 y 2 6 2 =1

Step 7: Identify h, k, a and b and the orientation of the transverse axis from the standard equation in Step 6.

Center: (h, k) = (-3, 0)


Note that x + 3 = x - (-3)


a = 2


b= 6


The transverse axis is horizontal since x is in the numerator above a2.

Step 8: Write the equations of the asymptotes

Equation for a horizontal transverse axis:


y=0± b a ( xk )


Substitute:


y=0± 6 2 [ x( 3 ) ]


Simplify:


y=± 6 2 ( x+3 )

Example 2: Find the standard equation of a hyperbola having vertices at (4, 3) and (4, 9) and asymptotes y=4±2x12 .

Step 1: Find the center coordinates.

Center: The center is the midpoint of the two vertices.


( h, k )=( 4+4 2 ,  3+9 2 )=( 8 2 , 12 2 )=( 4, 6 )

Step 2: Determine the orientation of the transverse axis and the distance between the center and the vertices (a).

Orientation of the transverse axis: Since both vertices fall on the vertical line x = 4, the transverse axis is vertical.


Length of a: To find the length between the center and the vertices take either vertices and find the change in their y-coordinates.


Vertex (4, 3): a=| 63 |=| 3 |=3


Vertex (4, 9): a=| 69 |=| 3 |=3


a = 3

Step 3: Determine the value of b.

The given asymptote equation, y=4±2x12 has a slope of 2. Because the transverse axis is vertical, 2=a/b .


b= a 2 = 3 2 b=1.5

Step 4: Write the standard form of the hyperbola.

Equation for a vertical transverse axis:


( yk ) 2 a 2 ( xh ) 2 b 2 =1


Substitute:


( y6 ) 2 3 2 ( x4 ) 2 ( 1.5 ) 2 =1


Final equation:


( y6 ) 2 3 2 ( x4 ) 2 ( 1.5 ) 2 =1





Related Links:
Math
algebra
Polar Coordinates: Coordinate Conversion
Polar Equation: Conversion Between Rectangular Form
Pre Calculus


To link to this Hyperbola: Asymptotes page, copy the following code to your site: