Hyperbola

Hyperbola
Helen M. Ogoc
Punta National High School
Punta, Dipolog City
With Teacher

Martin-Gay, Developmental Mathematics 2
Hyperbola
Definition:
A hyperbola is the set of all points P such that the
difference of the distance from P to two fixed points,
called the foci, is constant.
The line through the foci intersects the hyperbola at two
points, the vertices.
The line segment joining the vertices is the transverse
axis, and its midpoint is the center of the hyperbola.
A hyperbola has two branches and two asymptotes.

The asymptotes contain the diagonals of a
rectangle centered at the hyperbola’s center.
To get the correct shape of the hyperbola, we need
to find the asymptotes of the hyperbola.
The asymptotes are lines that are approached but
not touched or crossed.
These asymptotes are boundaries of the hyperbola.
This is one difference between a hyperbola and a
parabola.

For the hyperbolas that
open right/left,
the asymptotes are:
and for hyperbolas
opening up/down, the
asymptotes are:

Characteristics of Hyperbola (center of origin)
Equation Transverse Axis Asymptotes Vertices
Horizontal Y = ±(b/a)x (±a, 0)
Vertical Y = ±(a/b)x (0, ±a)
Foci: c² = a² + b²

Diagram of Hyperbola Standard Form

If the hyperbola opens right/left the translation is:
with the equations of the asymptotic lines as:
y - k = + (b/a)(x - h)
If the hyperbola opens up/down the translation is:
with the equations of the asymptotic lines as:
y - k = + (a/b)(x - h)

horizontal transverse axis
Hyperbola

Hyperbola horizontal transverse axis
"a" is the number in the denominator of the positive term
If the x-term is positive, then the hyperbola is horizontal
a = semi-transverse axis
b = semi-conjugate axis
center: (h, k)
vertices: (h + a, k), (h - a, k)

c = distance from the center to each focus along
the transverse axis
foci: (h + c, k), (h - c, k)
The eccentricity e > 1
The eccentricity e > 1

Draw the hyperbola given by
Solution:
Therefore, a = 3, b = 2 and c = 3.6.
This hyperbola opens right/left because it is in the form x - y.
a2
= 9, b2
= 4, c2
= 9 + 4 = 13.

a
b

Example 2:Example 2:

Solution: From the standard form:
x - h x - 2 h = 2
y - k y - 0 k = 0 Center: (2,0)
a2
= 4 a = 2 b2
= 25 b = 5
This means we move 2 units to the left and right of the
center and 5 units up and down from the center to arrive at
points on the guide rectangle.
The asymptotes pass through the center of the hyperbola as
well as the corners of the rectangle.

y2
term is being subtracted from the x2
term,
the branches of the hyperbola open to the left and right
this means that the transverse axis lies along the x-axis
the conjugate axis lies along the vertical line x = 2
the vertices of the hyperbola are where the hyperbola
intersects the transverse axis
the vertices are 2 units to the left and right of (2, 0)
which is the center
vertices: (0,0) and (4, 0)

To find the foci, we need
(-3.39,0) and (7.39,0)
To determine the equations of the asymptotes, recall that the
asymptotes go through the center of the hyperbola, (2; 0), as
well as the corners of guide rectangle,
Using the point-slope equation of a line,
We get,

Putting it all together, we get
Center: (2,0)
vertices: (0; 0) and (4; 0)
(-3.39,0) and (7.39,0)
Asymptotes:

Exercises :Exercises :
1. Graph the hyperbola. Find the center, the lines which
contain the transverse and conjugate axes, the vertices, the
foci and the equations of the asymptotes of
Center:
Tranverse Axis:
Congugate Axis:
Vertices:
Foci:
Asymptote:
(4, 2)
y = 2
x = 4

Center:
Tranverse Axis:
Congugate Axis:
Vertices:
Foci:
Asymptote:
(3,-1)
y = -1
x = 3
2. Put the equation in standard form. Find the center, the lines
which contain the transverse and conjugate axes, the vertices,
the foci and the equations of the asymptotes.
9x2
- 25y2
- 54x - 50y - 169 = 0
Equation in standard form:

3. Find the standard form of the equation of the hyperbola
which has the given properties..
a. Vertex (0, 1), Vertex (8, 1), Focus (-3, 1)
c. Vertices (3, 2), (13, 2);
Endpoints of the Conjugate Axis (8, 4), (8, 0)

Answers :Answers :
a. Vertex (0, 1), Vertex (8, 1), Focus (-3, 1)
c. Vertices (3, 2), (13, 2);
Endpoints of the Conjugate Axis (8, 4), (8, 0)

4. Find the standard form of the following equation using
then graph the conic section.
d. 9x2
- 4y2
- 36x - 24y - 36 = 0
a. x2
- 2x - 4y - 11 = 0
b. x2
+ y2
- 8x + 4y + 11 = 0
c. 9x2
+ 4y2
- 36x + 24y + 36 = 0

a. x2
- 2x - 4y - 11 = 0 b. x2
+ y2
- 8x + 4y + 11 = 0
Answers :Answers :

d. 9x2
- 4y2
- 36x - 24y - 36 = 0c. 9x2
+ 4y2
- 36x + 24y + 36 = 0

T
H
A
N K YO
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listening

Hyperbola

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