10.5 - Hyperbolas
10.5 - Hyperbolas
HYPERBOLA TERMS
c
a
C=(h , k)
Conjugate axis
Transverseaxis
EQUATION FORM
CENTER
VERTICES
CO-VERTICES
TRANSVERSE AXIS
TRANSVERSE length
CONJUGATE AXIS
CONJUGATE length
FOCI
ASYMPTOTES
(h, k )
(h, k ± b )
vertical
(h ± a , k)
b
Vertex
Co-vertex
Co-vertex
1b
)ky(
a
)hx(2
2
2
2
2a
horizontal
(h ± c , k)
222 bac
Vertex
2b
k)hx(a
by
Focus
The “Butterfly”
HYPERBOLA TERMS
c
a
C=(h , k)
Conjugate axis
Transverseaxis
EQUATION FORM
CENTER
VERTICES
CO-VERTICES
TRANSVERSE AXIS
TRANSVERSE length
CONJUGATE AXIS
CONJUGATE length
FOCI
ASYMPTOTES
(h, k )
(h ± b, k)
vertical
(h, k ± a )
b
Vertex
Co-vertex
Co-vertex
1b
)hx(
a
)ky(2
2
2
2
2a
horizontal
(h, k ± c )
222 bac
Vertex
2b
k)hx(b
ay
The “Hourglass”
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) = -19 + 64 – 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
14
)1y(
9
)4x( 22
Example Graph 4x2 – 16y2 = 64.
4x2 – 16y2 = 64
– = 1 Rewrite the equation in standard form. x 2
16y 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.
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.
ba
12
Step 4: Sketch the branches of the hyperbola through the vertices so they approach the asymptotes.
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 = 9 + 4 = 13• c² = 13 and c = ±√13• Focus points are √13 left and right
from the center F(-5 ±√13 , -2)
19
)2y(
4
)5x( 22
• a) GRAPH• Plot Center (-5,-2)• a = 2 (go left and right)• b = 3 (go up and down)
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 = 4 + 16 = 20• c² = 20 and c = ±2√5• Focus points are 2√5 up and down
from the center F(-1, 3 ±2√5)
116
)1x(
4
)3y( 22
• a) GRAPH• Plot Center (-1,3)• a = 2 (go up and down)• b = 4 (go left and right)
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
A = (-5,2) B = (5,2) major
conj
ugat
e
136
)2y(
25
)x( 22
1b
)ky(
a
)hx(2
2
2
2
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)
1144
)x(
25
)2y( 22
1b
)hx(
a
)ky(2
2
2
2
F(-3,2-13)
ExampleFind the foci of the graph – = 1.
x 2
9y 2
4
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.
= 9 + 4 Substitute 9 for a2 and 4 for b2.
c = 13 3.6 Find the square root of each side of the equation.
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.ba
23
Real Life ExamplesAs 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.
c2 = a2 + b2 Use the Pythagorean Theorem.
(424,650)2 = (300,765)2 + b2 Substitute.
1.803 1011 = 9.046 1010 + b2 Use a calculator.
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
Real Life Examples(continued)
b2 = 1.803 1011 – 9.046 1010 Solve for b2.
= 8.987 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