Hyperbola
See how to solve a hyperbola
(transverse axis, foci, equation, asymptotes).
6 examples and their solutions.
Definition
|PF - PF'| = (constant, transverse axis).
F, F': foci
Hyperbola: x2a2 - y2b2 = 1
Equation
x2a2 - y2b2 = 1
a2 + b2 = c2
Transverse Axis: 2a
Foci: (c, 0), (-c, 0)
a2 + b2 = c2
Transverse Axis: 2a
Foci: (c, 0), (-c, 0)
Example
x29 - y216 = 1
1. Transverse axis?
2. Foci?
Solution 1. Transverse axis?
2. Foci?
x29 - y216 = 1
x232 - y242 = 1
1. (transverse axis) = 2⋅3
= 6
2. c2 = 32 + 42
= 9 + 16
= 25
c = ±5
Foci: (5, 0), (-5, 0)
x232 - y242 = 1
1. (transverse axis) = 2⋅3
= 6
2. c2 = 32 + 42
= 9 + 16
= 25
c = ±5
Foci: (5, 0), (-5, 0)
Close
Example
Foci: (3, 0), (-3, 0)
Transverse axis: 4
Equation of the hyperbola?
Solution Transverse axis: 4
Equation of the hyperbola?
c = 3
2a = 4
a = 2
22 + b2 = 32
4 + b2 = 9
b2 = 5
x222 - y25 = 1 - [2]
x24 - y25 = 1
[1]
Draw the hyperbola
and the foci (3, 0), (-3, 0).
Foci are located horizontally.
So the transverse axis, 4, is horizontal.
So 2a = 4.
and the foci (3, 0), (-3, 0).
Foci are located horizontally.
So the transverse axis, 4, is horizontal.
So 2a = 4.
[2]
a = 2
b2 = 5
So the equation of the hyperbola is
x2/22 - y2/5 = 1.
b2 = 5
So the equation of the hyperbola is
x2/22 - y2/5 = 1.
Close
Asymptotes
x2a2 - y2b2 = 1
Asymptotes: y = ±bax
The graph of a hyperbola follows two asymptotes.Asymptotes: y = ±bax
(purple dashed lines)
Asymptotes: y = [b/a]x, y = -[b/a]x
Example
x29 - y216 = 1
Asymptotes?
Solution Asymptotes?
x29 - y216 = 1
x232 - y242 = 1
Asymptotes: y = ±43x
x232 - y242 = 1
Asymptotes: y = ±43x
Close
Hyperbola: x2a2 - y2b2 = -1
Equation
x2a2 - y2b2 = 1
a2 + b2 = c2
Transverse Axis: 2b
Foci: (0, c), (0, -c)
a2 + b2 = c2
Transverse Axis: 2b
Foci: (0, c), (0, -c)
Example
4x2 - y2 = -4
1. Transverse axis?
2. Foci?
Solution 1. Transverse axis?
2. Foci?
4x2 - y2 = -4
x21 - y24 = -1
x212 - y222 = -1
1. (transverse axis) = 2⋅2
= 4
2. c2 = 12 + 22
= 1 + 4
= 5
c = ±√5
Foci: (√5, 0), (-√5, 0)
x21 - y24 = -1
x212 - y222 = -1
1. (transverse axis) = 2⋅2
= 4
2. c2 = 12 + 22
= 1 + 4
= 5
c = ±√5
Foci: (√5, 0), (-√5, 0)
Close
Example
Foci: (0, 5), (0, -5)
Transverse axis: 6
Equation of the hyperbola?
Solution Transverse axis: 6
Equation of the hyperbola?
c = 5
2b = 6
b = 3
a2 + 32 = 52
a2 + 9 = 25
a2 = 16
x216 - y232 = -1 - [2]
x216 - y29 = -1
[1]
Draw the hyperbola
and the foci (0, 5), (0, -5).
Foci are located vertically.
So the transverse axis, 6, is vertical.
So 2b = 6.
and the foci (0, 5), (0, -5).
Foci are located vertically.
So the transverse axis, 6, is vertical.
So 2b = 6.
[2]
a2 = 16
b = 3
So the equation of the hyperbola is
x2/16 - y2/32 = -1.
b = 3
So the equation of the hyperbola is
x2/16 - y2/32 = -1.
Close
Asymptotes
x2a2 - y2b2 = -1
Asymptotes: y = ±bax
The asymptotes formula is the sameAsymptotes: y = ±bax
as the upper hyperbola.
Asymptotes: y = [b/a]x, y = -[b/a]x
Example
y2 - 9x2 = 9
Asymptotes?
Solution Asymptotes?
y2 - 9x2 = 9
-9x2 + y2 = 9
x21 - y29 = -1
x212 - y232 = -1
Asymptotes: y = ±31x
y = ±3x
-9x2 + y2 = 9
x21 - y29 = -1
x212 - y232 = -1
Asymptotes: y = ±31x
y = ±3x
Close