1. LECTURE UNIT 006
Hyperbola
Set of all moving points in a plane such that the difference of the distance of each from two fixed points
is the same constant. The fixed points are called the and the line through them is the .
y
Directrix
Directrix
d4
d2 d1
Conjugate axis, 2b
d3
Where:
Latus Rectum
b
F2 v2 v1 F1
x F1, F2 = Focus
C (h, k)
b v1, v2 = Vertex
C = Center
d2 - d1 = 2a
d d
a a
Transverse axis, 2a
c c
Asymptote
y - k = -m (x - h)
Elements of Hyperbola
c2 = a2 + b2
d3 c
Eccentricity, e = = a >1
d4
a
d= e
2b2
Latus rectum, LR = a
Equation of Asymptote
Asymptote of a curve is a line when the perpendicular distance from a line to a curve
approaches zero as the curve extends indefinitely far from the origin.
y - k =+ m (x - h)
-
Where (h, k) is the center of the hyperbola and m is the slope. m = b/a if the axis is
horizontal and m = a/b if the axis is vertical. Use (+) for upward asymptote and (-) for
downward asymptote.
Standard Equations
Where the transverse axis is parallel to x-axis (opens left and right)
(x - h)2 (y - k)2
- =1
a2 b2
Where the transverse axis is parallel to y-axis (opens up and down)
(y - k)2 (x - h)2
- =1
a2 b2
Length of transverse axis (T.A.) = 2a
Length of conjugate axis (C.A.) = 2b
Length of focal axis (F.A.) = 2c
“Some people succeed because they are destined to, but most people succeed
because they are determined.”

2. Sketch the graph:
(x - 2)2 (y + 3)2
1. - =1
4 9
(y - 1)2 (x + 3)2
2. - =1
16 9
The equation Ax2 + By2+ Dx + Ey + F = 0 where A, B < 0 is the general equation of the hyperbola. To sketch
the graph, reduce the equation to standard form.
2 2
D E
(
A x + 2A ) -B (y + 2B ) =M
Note:
M = 0 consist of two intersecting lines
M = 0 graph is a hyperbola
3. 4x2 - 9y2 + 16x - 18y - 29 = 0
4. 5x2 - 3y2 + 10x - 6y + 2 = 0
5. 16x2 - 25y2 + 64x + 100y + 364 = 0
Find the equation of the hyperbola with given conditions.
6. With foci at (-2, 5) and (-2, -5) and a vertex at (-2, 4).
7. With center at the origin and passing through (-2, 3) and (1, -1).
8. With center at the intersection of x + 2y = 2 and 3x - 2y = 2, a vertex at (4, ½) and length of the
latus rectum is 9/2.
9. With vertices at (-3, 2) and (5, 2) and length of the focal axis is the diameter of the circle x2 + y2 - 10x = 0.
10. With center at the vertex of the parabola x2 - 2y + 4x = 0, foci at an end of conjugate axis at
(-4, -2) and (-6, -2).
“A leader is an individual who has an inspiring vision and can get others to buy
into it.”

3. HYPERBOLA
Example 1:
(x - 2)2 (y + 3)2
- =1
4 9
T.A. parallel to the x-axis
C (2, -3)
C(2, -3)
a2 = 4 b2 = 9
a=2 b=3
2b2
LR = a =9
c2 = a2 + b2
c = 3.6
Example 5:
16x2 - 25y2 + 64x + 100y + 364 = 0
(16x2 + 64x) + (-25y2 + 100y) = -364
16(x2 + 4x + 4) - 25(y2 - 4y + 4) = -364 + 64 - 100
16(x + 2)2 - 25(y - 2)2 = -400
(y - 2)2 (x + 2)2
- =1
16 25
T.A. parallel to the y-axis
C (-2, 2)
a2 = 16 b2 = 25
a=4 b=5
Example 7:
With center at the origin and passing through (-2, 3) and (1, -1).
From the standard equation:
x2 y2
2
- 2 =1
a b
At (-2, 3)
4 9
2
- 2 =1
a b
At (1, -1)
1 1
2
- 2 =1
a b
2 5 and a2 = 5
By elimination and substitution, we obtain: b =
3 8
Substituting;
x2 y2
- =1
5 5
8 3
Simplifying, the equation of the hyperbola is
8x2 - 3y2 = 5
“Treat people as if they were what they ought to be, and you may help them to
become what they are capable of being.”