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