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10.5 - Hyperbolas

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10.5 - Hyperbolas

Vertex

The “Butterfly”

(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)

Vertex

The “Hourglass”

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 )

- 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

4x2 – 16y2 = 64

– = 1Rewrite the equation in standard form.

The equation of the form – - = 1, so the transverse axis is horizontal.

Since a2 = 16 and b2 = 4, a = 4 and b = 2.

y2

y2

x2

x2

16

a2

4

b2

Graph 4x2 – 16y2 = 64.

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.

b

a

1

2

Step 4:Sketch the branches of the hyperbola through the vertices so they approach the asymptotes.

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.

- a) GRAPH
- Plot Center (-5,-2)
- a = 2 (go left and right)
- b = 3 (go up and down)

- 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)

- a) GRAPH
- Plot Center (-1,3)
- a = 2 (go up and down)
- b = 4 (go left and right)

- 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)

- 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

F(-3,2+13)

- 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)

y2

The equation is in the form – - = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4.

y2

x2

b2

9

4

c2 = a2 + b2Use the Pythagorean Theorem.

= 9 + 4Substitute 9 for a2 and 4 for b2.

x2

a2

c = 13 3.6Find the square root of each side of the equation.

Find the foci of the graph – = 1.

b

a

2

3

The asymptotes are the lines y = ± x , or y = ± x.

(continued)

The foci (0, ±c) are approximately (0, –3.6) and (0, 3.6). The vertices (0, ±b) are (0, –2) and (0, 2).

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.

x2

a2

c2 = a2 + b2Use the Pythagorean Theorem.

y2

b2

(424,650)2 = (300,765)2 + b2Substitute.

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.

1.803 1011 = 9.046 1010 + b2Use a calculator.

b2 = 1.803 1011 – 9.046 1010Solve for b2.

= 8.987 1010

– = 1Substitute a2 and b2.

– = 1.

The path of the spacecraft around the planet can be modeled by

x2

x2

y2

y2

9.046 1010

9.046 1010

8.987 1010

8.987 1010

(continued)