1. Hyperbolas

2. HYPERBOLA TERMS Vertex The “Butterfly” EQUATION FORM CENTER VERTICES
CO-VERTICES
TRANSVERSE AXIS
TRANSVERSE length
CONJUGATE AXIS
CONJUGATE length
FOCI
ASYMPTOTES
(h, k )
Conjugate axis
(h ± a , k)
Co-vertex
(h, k ± b )
Vertex
Focus b
horizontal a 2a c
vertical
Transverse
axis
Vertex
C=(h , k) 2b
Co-vertex
(h ± c , k)

3. HYPERBOLA TERMS Vertex The “Hourglass” EQUATION FORM CENTER VERTICES
CO-VERTICES
TRANSVERSE AXIS
TRANSVERSE length
CONJUGATE AXIS
CONJUGATE length
FOCI
ASYMPTOTES
Transverse
axis
(h, k )
Vertex
(h, k ± a )
(h ± b, k)
C=(h , k)
Co-vertex
vertical c 2a a b
horizontal
Co-vertex
Conjugate
axis 2b
Vertex
(h, k ± c )

4. CONVERTING to STANDARD FORM
4x² – 9y² – 32x – 18y + 19 = 0
Groups the x terms and y terms
4x² – 32x – 9y²– 18y + 19 = 0
Complete the square
4(x² – 8x) – 9(y² + 2y) + 19 = 0
4(x² – 8x + 16) – 9(y² + 2y + 1) = – 9
4(x – 4)² – 9(y + 1)² = 36
Divide to put in standard form
4(x – 4)²/36 – 9(y + 1)²/36 = 1

5. Example Graph 4x2 – 16y2 = 64. 4x2 – 16y2 = 64
– = 1 Rewrite the equation in standard form.
x 2 16
y 2 4
Since a2 = 16 and b2 = 4, a = 4 and b = 2.
The equation of the form – = 1, so the transverse axis is horizontal.
x 2
a 2
y 2
b 2
Step 1: Graph the vertices. Since the transverse axis is horizontal, the vertices lie on the x-axis. The coordinates are (±a, 0), or (±4, 0).
Step 2: Use the values a and b to draw the central “invisible” rectangle. The lengths of its sides are 2a and 2b, or 8 and 4.

6. Example
Graph 4x2 – 16y2 = 64.
Step 3: Draw the asymptotes. The equations of the asymptotes are y = ± x or y = ± x . The asymptotes contain the diagonals of the central rectangle. b a 1 2
Step 4: Sketch the branches of the hyperbola through the vertices so they approach the asymptotes.

7. Graph and Label b) Find coordinates of vertices, covertices, foci
Center = (-5,-2)
Butterfly shape since the x terms come first
Since a = 2 and b = 3
Vertices are 2 points left and right from center  (-5 ± 2, -2)
CoVertices are 3 points up and down  (-5, -2 ± 3)
Now to find focus points
Use c² = a² + b²
So c² = = 13
c² = 13 and c = ±√13
Focus points are √13 left and right from the center  F(-5 ±√13 , -2)
a) GRAPH
Plot Center (-5,-2)
a = 2 (go left and right)
b = 3 (go up and down)

8. Graph and Label b) Find coordinates of vertices, covertices, foci
Center = (-1,3)
Hourglass shape since the y terms come first
Since a = 2 and b = 4
Vertices are 2 points up and down from center  (-1, 3 ± 2)
Covertices are 3 points left and right  (-1 ± 4, 3)
Now to find focus points
Use c² = a² + b²
So c² = = 20
c² = 20 and c = ±2√5
Focus points are 2√5 up and down from the center  F(-1, 3 ±2√5)
a) GRAPH
Plot Center (-1,3)
a = 2 (go up and down)
b = 4 (go left and right)

9. Write the equation of the hyperbola given… vertices are at (-5,2) and (5,2) conjugate axis of length 12
Draw a graph with given info
Use given info to get measurement
Find the center first
Center is in middle of vertices,
so (h , k) = (0 , 2)
A = distance from center to vertices,
so a = 5
Also, the conjugate length = 2b
Since conjugate = 12
Then b = 6
Use standard form
Need values for h,k, a and b
We know a = 5 and b = 6
The center is (0, 2)
Plug into formula
B = (5,2)
A = (-5,2)
conjugate
major

10. Write the equation of the hyperbola given… center is at (-3,2) foci at (-3, 2±13) and major axis is 10
Draw a graph with given info
Use given info to get measurements
Find the center first
Center is in middle of vertices,
so (h , k) = (0 , 2)
A = distance from center to vertices,
so a = 5
We still don’t have b ….
Use the formula  c² = a² + b²
Since a = 5 and c = 13 then….
b = 12 (pythagorean triplet)
Use standard form
Need values for h,k, a and b
We know a = 5 and b = 12
and center is (0, 2)
Plug into formula
F(-3,2+13)
F(-3,2-13)

11. Example Find the foci of the graph – = 1.
y 2 4
Find the foci of the graph – = 1.
x 2 9
The equation is in the form – = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
= Substitute 9 for a2 and 4 for b2.
c = Find the square root of each side of the equation.

12. Example
(continued)
The foci (0, ±c) are approximately (0, –3.6) and (0, 3.6). The vertices (0, ±b) are (0, –2) and (0, 2).
The asymptotes are the lines y = ± x , or y = ± x. b a 2 3

13. Real Life Examples
As a spacecraft approaches a planet, the gravitational pull of the planet changes the spacecraft’s path to a hyperbola that diverges from its asymptote. Find an equation that models the path of the spacecraft around the planet given that a = 300,765 km and
c = 424,650 km.
Assume that the center of the hyperbola is at the origin and that the transverse axis is horizontal. The equation will be in the form – = 1.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
(424,650)2 = (300,765)2 + b2 Substitute.
1.803  1011 =  b2 Use a calculator.

14. Real Life Examples b2 = 1.803  1011 – 9.046  1010 Solve for b2.
(continued)
b2 =  1011 –  1010 Solve for b2.
=  1010
– = 1 Substitute a2 and b2.
x 2
9.046  1010
y 2
8.987  1010
The path of the spacecraft around the planet can be modeled by
– = 1.
x 2
9.046  1010
y 2
8.987  1010