1. The Hyperbola
Week 18

2. Definition of a hyperbola
The locus of a point which moves in such a way that the ratio of its distance from a fixed point (focus) and from a fixed straight line (directrix) is a constant greater than 1 is a hyperbola.

3. The Equation of a Hyperbola
S is the focus
ZL is the directrix
A is a point on the hyperbola
O is the mid-point of AA’
SA = e.AZ (e>1)
P is another point on the hyperbola L N
P(x, y) A’ O Z A S

4. The Equation of a Hyperbola
SA= e.AZ (e>1)
SA’=e.A’Z
SA’-SA=e(A’Z-AZ)
(OS+a)-(OS-a)=e(A’Z-AZ)
2a=e[(a+OZ)-(a-OZ)]
2a=2e.OZ
OZ=a/e (Important) L N
P(x, y) A’ O Z A S

5. The Equation of a Hyperbola
(OS-a)=SA
(OS-a)=eAZ
OS-a=e(a-OZ)
OS=a+e(a-a/e)
OS=a+ae-a
OS=ae (Important) L N
P(x, y) A’ O Z A S

6. The Equation of a Hyperbola
P(x, y) A’ O Z A S
PS = e.PN (e>1)
PS2=e2PN

7. Alternative Definition
The set of all points in a plane such that the difference of the distances to 2 fixed points (foci) is constant.
Standard form of a hyperbola centered at the origin:
Opens left & right
Opens up & down

8. What are Hyperbolas? Hyperbolas can be thought of as two parabolas going in opposite directions

9. Hyperbola F1 F2 d1 d2 P
For any point P that is on the hyperbola, d2 – d1 is always the same (is constant).
In this example, the origin is the center of the hyperbola. It is midway between the foci.

10. Hyperbolas F V C
1. A line through the foci intersects the hyperbola at two points, called the vertices (V)
2. The segment connecting the vertices is called the transverse axis of the hyperbola.
3. The center of the hyperbola is located at the midpoint of the transverse axis.

11. Hyperbolas F V C
As x and y get larger the branches of the hyperbola approach a pair of intersecting lines called the asymptotes of the hyperbola.
These asymptotes pass through the center of the hyperbola.

12. The Standard Equation of a Hyperbola
With Center (0, 0) and Foci on the x-axis
The equation of a hyperbola
with the centre (0, 0) and
foci on the x-axis is:
B (0, b)
The length of the transverse
axis is 2a.
The length of the conjugate
axis is 2b.
The vertices are (a, 0) and
(-a, 0).
The foci are (c, 0) and (-c, 0).
The slopes of the asymptotes are
(-c, 0) A1 A2
(c, 0) F1
(-a, 0)
(a, 0) F2
B (0, -b)
The equations of the asymptotes

13. The Standard Equation of a Hyperbola with
Center (0, 0) and Foci on the y-axis
The equation of a hyperbola
with the centre (0, 0) and
foci on the y-axis is:
F1(0, c)
The length of the transverse
axis is 2a.
The length of the conjugate
axis is 2b.
The vertices are (0, a) and
( 0, -a).
The foci are (0, c) and (0, -c).
The slopes of the asymptotes are
A1(0, a)
B1(-b, 0)
B2(b, 0)
A2(0, -a)
F2(0, -c)
The equations of the asymptotes

14. State the coordinates of the vertices, the coordinates of the foci,
Hyperbola Properties
State the coordinates of the vertices, the coordinates of the foci,
the lengths of the transverse and conjugate axes, and the equations
of the asymptotes of the hyperbola defined by each equation.
The equations of the asymptotes are
For this equation, a = 2 and b = 4.
The length of the transverse axis is
2a = 4.
The length of the conjugate axis is
2b = 8.
The vertices are (2, 0) and (-2, 0):
c2 = a2 + b2 =
= 20
The coordinates of the foci are

15. Hyperbola Properties For this equation, a = 5 and b = 3.
The length of the transverse axis is
2a = 10.
The length of the conjugate axis is
2b = 6.
The vertices are (0, 5) and (0, -5):
c2 = a2 + b2 =
= 34
The coordinates of the foci are
The equations of the asymptotes are

16. The Standard Form of the Hyperbola with Centre (h, k)
When the transverse axis is vertical,
the equation in standard form is:
The centre is (h, k).
The transverse axis is parallel to the
y-axis and has a length of 2a units.
The conjugate axis is parallel to the
x-axis and has a length of 2b units.
The slopes of the asymptotes are
(h, k)
The general form of the equation is
Ax2 + Cy2 + Dx + Ey + F = 0.

17. The Standard Form of the Hyperbola with Centre (h, k) [cont’d]
When the transverse axis is
horizontal, the equation in
standard form is:
The transverse axis is parallel to the
x-axis and has a length of 2a units.
The conjugate axis is parallel to the
y-axis and has a length of 2b units.
The slopes of the asymptotes are

18. Describe the hyperbola
Transverse axis is vertical
Centered at (-1,3)
Distance to vertices from center = 2 units (up & down) (-1,5) & (-1,1)
Asymptotes pass through the (-1,3) with slopes = 2/3, -2/3
Foci units up & down from the center ,

19. Finding the Equation of a Hyperbola
The centre is (2, 3), so h = 3 and k = 2.
The transverse axis is parallel to the
y-axis and has a length of 10 units, so a = 5.
The conjugate axis is parallel to the
x-axis and has a length of 6 units, so b = 3.
The vertices are (-2, 8) and (-2, -2).
The slope of one asymptote is ,
so a = 5 and b = 3:
c2 = a2 + b2 =
= 34
The coordinates of the foci are
Standard
form

20. Writing the Equation in its General Form
9(y - 3)2 - 25(x +2)2 = 225
9(y2 - 6y + 9) - 25(x2 + 4x + 4) = 225
9y2 - 54y x x = 225
-25x2 + 9y x - 54y = 225
-25x2 + 9y x - 54y = 0
The general form of the equation

21. Writing the Equation of a Hyperbola
Write the equation of the hyperbola with centre at (2, -3), one
vertex at (6, -3), and the coordinates of one focus at (-3, -3).
The centre is (2, -3), so h = 2, k = -3.
The distance from the centre to the vertex is 4 units, so a = 4.
The distance from the centre to the foci is 5 units, so c = 5.
Use the Pythagorean property to find b:
b2 = c2 - a2 =
= 9
b = ± 3
9(x - 2)2 - 16(y + 3)2 = 1
9(x2 - 4x + 4) - 16(y2 + 6y + 9) = 144
9x2 - 36x y2 - 96y = 144
9x2 - 16y2 - 36x - 96y = 144
9x2 - 16y2 - 36x - 96y = 0
General form
Standard
form

22. State the coordinates of the vertices, the coordinates of
Sample Problem
State the coordinates of the vertices, the coordinates of
the foci, the lengths of the transverse and conjugate axes
and the equations of the asymptotes of the hyperbola
defined by 4x2 - 9y2 + 32x + 18y + 91 = 0.
4x2 - 9y2 + 32x + 18y + 91 = 0
(4x2 + 32x ) + (- 9y2 + 18y) + 91 = 0
4(x2 + 8x + ____) - 9(y2 - 2y + _____) = _____ + _____ 16 1 64 -9
4(x + 4)2 - 9(y - 1)2 = -36
3.5.14

23. General Information about Hyperbolas Centered at Origin
x and y are both squared
Equation always equals(=) 1
Equation is always minus(-)
a2 is always the first denominator
c2 = a2 + b2
c is the distance from the center to each foci on the major axis
a is the distance from the center to each vertex on the major axis

24. More General Information about Hyperbolas Centered at the Origin
b is the distance from the center to each midpoint of the rectangle used to draw the asymptotes. This distance runs perpendicular to the distance (a).
Major axis has a length of 2a
Eccentricity(e): e = c/a (The closer e gets to 1, the closer it is to being circular
If x2 is first then the hyperbola is horizontal
If y2 is first then the hyperbola is vertical.

25. What changes if the center is NOT at the origin?
Standard Form of Equation
Equations of Asymptotes

26. Essential Properties of Hyperbolas
Equation of Hyperbola
Center
Foci Points
Equation of Asymptote
Vertex
Transverse Axis
(x – h)2 _ (y – k)2
a b2
(h, k)
(h – c, k) and (h + c, k)
y – k = +/- (b/a) (x – h
(h +a, k) and (h – a, k)
Horizontal
(y – k)2 _ (x – h)2
a b2
(h, k – c) and (h, k + c)
(c = a2 + b2 )
y – k = +/- (a/b) (x – h)
(h, k + a) and
(h, k – a)
Vertical
= 1
= 1
You must be looking at a hyperbola when the x2 and y2 terms are subtracted (x2 – y2) or (y2 – x2)

27. Practice Example 1
Write the standard form of the equation of the hyperbola
144y2 – 25x2 – 576y – 150x = 3249.
Then find the coordinates of the center, the vertices, the foci, and the equation of the asymptotes. Graph the hyperbola and the asymptotes.

28. Practice Example 1
Write the standard form of the equation of the hyperbola y2 – 25x2 – 576y – 150x = Then find the coordinates of the center, the vertices, the foci, and the equation of the asymptotes. Graph the hyperbola and the asymptotes.
144(y2 – 4y + o) – 25(x2 + 6x + o) = (o) + 25(o)
144(y2 – 4y + 4) – 25(x2 + 6x + 9) = (4) + 25(9)
144(y – 2)2 – 25(x + 3)2 = 3600
(y-2)2 _ (x + 3)2
= 1
Center: (-3, 2) a = 5 so the vertices are (-3, 7) and (-3, -3)
a2 + b2 = c2
= c2
c = 13
The foci are (-3, 15) and (-3, -11).

29. Practice Example 1 (cont.)
Asymptotes have the formula y = +/- a/b x and we have center (-3, 2) and slopes +/- 5/12.
y – 2 = 5/12 (x + 3) y – 2 = -5/12 (x + 3)
y – 2 = (5/12) x + 15/12 y – 2 = (-5/12) x + -15/12
y = (5/12) x + 13/4 y = (-5/12) x + 3/4

30. Practice Problem 2
Find the coordinates of the vertices and the foci. Give the asymptote slopes for each hyperbola. Then draw the graph.
x2 _ y2
2) 25x2 – 4y2 = 100
= 1
Vertices: (-2, 0) (2, 0) Foci: ( 29, 0) (- 29, 0) Slope = +/- 5/2
Vertices: (-3, 0) (3, 0) Foci: (- 58, 0) ( 58, 0) Slope = +/- 7/3 2) 1)

31. Further Problems
Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four conic sections. Then graph the conic section.
9x2 + 16y2 – 54x + 64y = -1
9 (x2 – 6x + o) + 16(y2 + 4y + o) = (o) + 16(o)
9 (x2 – 6x + 9) + 16(y2 + 4y + 4) = (9) + 16(4)
9(x – 3)2 + 16(y + 2)2 = 144
(x – 3) (y + 2)2 9
This conic section is an ellipse. +
= 1

32. Further Problems
Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four conic sections. Then graph the conic section.

33. 1) (x + 3)2 + (y)2 = 16 circle 2) (x – 3)2 _ (y + 1)2 hyperbola
Further Problems
Write the equation in standard form and decide if the conic section is a parabola, a circle, an ellipse, or a hyperbola. Then graph the equation.
x2 + y2 + 6x = 7
5x2 – 6y2 – 30x – 12y + 9 = 0
1) (x + 3)2 + (y)2 = 16 circle ) (x – 3)2 _ (y + 1) hyperbola
= 1 2) 1)