The document discusses hyperbolas, which are conic sections that have two disconnected branches. Key elements of a hyperbola include its foci, vertices, transverse/conjugate axes, center, and asymptotes. The eccentricity measures the "bendiness" of the hyperbola. Equations of hyperbolas depend on whether the transverse axis is horizontal or vertical. The document provides examples and exercises on finding equations and properties of hyperbolas given certain information.
2. Hyperbola
• A conic section that can be thought of as
an inside-out ellipse.
• A hyperbola is the locus of points where
the difference in the distance to two fixed
points (called the foci) is constant.
3. • A hyperbola is the set of all points (x, y)
in the plane the difference of whose
distances from two fixed points is some
constant. The two fixed points are called
the foci.
• A hyperbola comprises two disconnected
curves called its arms or branches which
separate the foci.
6. • Foci
The foci are the fixed points of the hyperbola.
They are denoted by F and F'.
• Transverse Axis or real axis
The tranverse axis is the line segment between
the foci.
• Conjugate Axis or imaginary axis
The conjuagate axis is the perpendicular bisector
of the line segment (transverse axis).
• Center
The center is the point of intersection of the axes
and is also the center of symmetry of the
hyperbola.
7. • Vertices
The points A and A' are the points of intersection
of the hyperbola with the transverse axis.
• Focal Radii
The focal radii are the line segments that join a
point on the hyperbola with the foci: PF and PF'.
• Focal Length
The focal length is the line segment , which has
a length of 2c.
8. • Semi-Major Axis
The semi-major axis is the line segment that runs
from the center to a vertex of the hyperbola. Its
length is a.
• Semi-Minor Axis
The semi-minor axis is a line segment which is
perpendicular to the semi-major axis. Its length
is b.
• Axes of Symmetry
The axes of symmetry are the lines that coincide
with the transversal and conjugate axis.
9. • Asymptotes
The asymptotes are the lines with the
equations:
• The Relationship Between the Semi axes:
• Eccentricity of a Hyperbola:
10. Hyperbola is actually two separate curves in
mirror image.
On the diagram you can
see:
•a directrix and a focus (one
on each side)
•an axis of symmetry or
transverse axis (that goes
through each focus, at
right angles to the
directrix)
•two vertices (where each
curve makes its sharpest
turn)
11. Equation
• By placing a hyperbola on an x-y graph
(centered over the x-axis and y-axis), the
equation of the curve is:
x2/a2 - y2/b2 = 1
12. Also:
One vertex is at (a, 0),
and the other is at (-a,
0)
The asymptotes are the
straight lines:
•y = (b/a)x
•y = -(b/a)x
And the equation is also
similar to the equation
of the ellipse: x2/a2+
y2/b2 = 1, except for a
"-" instead of a "+")
13. When the transverse axis is
horizontal (in other words,
when the center, foci, and
vertices line up side by side,
parallel to the x-axis), then
the a2 goes with the x part of
the hyperbola's equation, and
the y part is subtracted.
14. When the transverse axis is
vertical (in other words,
when the center, foci, and
vertices line up above and
below each other, parallel to
the y-axis), then the a2 goes
with the y part of the
hyperbola's equation, and
the x part is subtracted.
15. Asymptotes
The graph of a hyperbola gets fairly flat and
straight when it gets far away from its center. If
you "zoom out" from the graph, it will look very
much like an "X", with a little curviness near the
middle. These "nearly straight" parts get very
close to what are called the "asymptotes" of the
hyperbola. For a hyperbola centered at (h, k) and
having fixed values a and b, the asymptotes are
given by the following equations:
17. Note that the only difference
in the asymptote equations is
in the slopes of the straight
lines:
If a2is the denominator for
the x part of the hyperbola's
equation, then a is still in the
denominator in the slope of
the asymptotes' equations.
18. If a2 goes with the y part of
the hyperbola's equation,
then a goes in the
numerator of the slope in
the asymptotes' equations.
19.
20.
21. Eccentricity
• The measure of the amount of curvature is the
"eccentricity" e, where e = c/a. Since the foci are
further from the center of an hyperbola than are
the vertices (so c > a for hyperbolas), then e >
1. Bigger values of e correspond to the
"straighter" types of hyperbolas, while values
closer to 1 correspond to hyperbolas whose
graphs curve quickly away from their centers.
22. On this diagram:
• P is a point on the curve,
• F is the focus and
• N is the point on the
directrix so that PN is
perpendicular to the
directrix.
The ratio PF/PN is the eccentricity of the hyperbola
(for a hyperbola the eccentricity is always greater than 1).
It can also given by the formula: e =
23. Hyperbolas can be fairly "straight" or else
pretty "bendy":
Hyperbola with an
eccentricity of about
1.05
Hyperbola with an
eccentricity of about 7.6
24. Latus Rectum
The Latus Rectum is the
line through the focus
and parallel to the
directrix.
The length of the Latus
Rectum is 2b2/a.
25. Hyperbola on the Horizontal Axis
Center at the Origin
Equation: x2/a2 - y2/b2 = 1
C (0,0)
V (± a,0)
F (±c,0)
Length of LR = 2b2/a
Ends of LR (c,±b2/a) and (-c,±b2/a)
Ends of Conjugate Axis (0,±b)
Asymptotes y = ± (b/a)x
e = c/a
c2 = a2+b2
26. Hyperbola on the Horizontal Axis
Center at (h,k)
Equation: (x-h)2/a2 - (y-k)2/b2 = 1
C (h,k)
V (h± a,k)
F (h±c,k)
Length of LR = 2b2/a
Ends of LR (h-c,k±b2/a) and (h-c,k±b2/a)
Ends of Conjugate Axis (h,k±b)
Asymptotes y = ± (b/a)x
e = c/a
c2 = a2+b2
27. Hyperbola on the Vertical Axis
Center on the Origin
Equation: y2/a2 - x2/b2 = 1
C (0,0)
V (0,± a)
F (0,±c)
Length of LR = 2b2/a
Ends of LR (±b2/a,c) and (±b2/a,-c)
Ends of Conjugate Axis (±b,0)
Asymptotes y = ± (a/b)x
e = c/a
c2 = a2+b2
28. Hyperbola on the Vertical Axis
Center at (h,k)
Equation: (y-k)2/a2 - (x-h)2/b2 = 1
C (h,k)
V (h,k± a)
F (h,k±c)
Length of LR = 2b2/a
Ends of LR (h±b2/a,k+c) and (h±b2/a,k-c)
Ends of Conjugate Axis (h±b,k)
Asymptotes y = ± (a/b)x
e = c/a
c2 = a2+b2
29. Exercises
1.) Find the center, vertices, foci, eccentricity, and
asymptotes of the hyperbola with the given
equation, and sketch:
.
2.) Give the center, vertices, foci, and asymptotes
for the hyperbola with equation:
.
30. 3.) Find the center, vertices, and asymptotes of the
hyperbola with equation
4x2 – 5y2 + 40x – 30y – 45 = 0.
4.) Find an equation for the hyperbola with
center (2, 3), vertex (0, 3), and focus (5, 3).
5.) Find an equation for the hyperbola with
center (0, 0), vertex (0, 5), and asymptotes y =
± (5/3)x.
31. 6.) Find an equation of the hyperbola with xintercepts at x = –5 and x = 3, and foci at (–6,
0) and (4, 0).
7.) Find an equation for the hyperbola with
vertices at (–2, 15) and (–2, –1), and having
eccentricity e = 17/8.
8.) Give the equation of the hyperbola with
vertices at (-7,0) and (7,0) and conjugate axis
length of 10.
32. Answers:
1.)
c2 = a2 + b2
c2 = 144 + 25 = 169
c = 13
e = c/a = 13/5
Since x2 = (x – 0)2 and y2 = (y – 0)2, then the
center is at (h, k) = (0, 0).
F (0,-13) and (0,13)
V (0,5) and (0,-5)
33. center (0, 0), vertices (0, –5) and
(0, 5), foci (0, –13) and (0, 13),
eccentricity
, and
asymptotes
34. 2.)
a2 = 16 and b2 = 9, so a = 4 and b = 3.
C = (h, k) = (–3, 2)
c2 = 9 + 16 = 25
c=5
F = (h-c, k)
= (-3-5, 2)
= (-8, 2)
Asymptotes
= (h+c, k)
= (-3+5, 2)
= (2, 2)
V = (h-a, k)
= (-3-4, 2)
= (-7, 2)
y – 2 = ± (3/4)(x + 3).
= (h+a, k)
= (-3+4, 2)
= (1,2)
36. C = (h, k) = (–5, –3)
a2 = 25
b2 = 20
a=5
b = 2sqrt[5]
c2 = 25 + 20 = 45
c =sqrt[45] = 3sqrt[5]
Asymptotes = m = ± (2/5)sqrt[5]
center (–5, –3), vertices (–10, –3) and (0, –3),
foci
and
, and asymptotes
37. 4.) C (2, 3), V (0, 3), and F (5, 3)
V (h+a, k)
2-a = 0
-a = -2
a=2
Eq.
F (h+c, k)
2+c = 5
c = 5-2
c=3
38. 5.) C (0, 0), V (0, 5), and asymptotes y = ± (5/3)x
V (h, k+a)
k+a = 5
0+a = 5
a=5
a2 = 25
Eq.
asymptotes = a/b
= 5/3
= 5/b
b=3
b2 = 9
39. 6.) x-intercepts at x = –5 and x = 3, and foci at
(–6, 0) and (4, 0)
C (h, k) = (–1, 0)
The foci are 5 units to either side of the center,
so c = 5 and c2 = 25
The center lies on the x-axis, so the two xintercepts must then also be the hyperbola's
vertices. Since the intercepts are 4 units to either
side of the center, then a = 4 and a2 = 16.
a2 +b2 = c2
b2 = 25 – 16 = 9
Eq.
40. 7.) V (–2, 15) and (–2, –1), and e = 17/8
(h, k) = (–2, 7)
The vertices are 8 units above and below the
center, so a = 8 and a2 = 64.
e = c/a
a2 + b2 = c2
=17/8
b2 = 289 – 64 = 225
c = 17
c2 = 289
Eq.
41. 8.) V (-7,0) (7,0) length of conjugate axis = 10
The hyperbola has its center at the origin
because the center is halfway between the
vertices and the origin is halfway between the
vertices. The equation of a hyperbola which
opens right and left is where a = half the length
of the transverse axis (the transverse axis is the
line between the vertices)
and b = half the length of the conjugate axis
(the conjugate axis is a line perpendicular to the
transverse axis at the center, half of which is
above the center and half below.
42. So since the vertices are (-7,0) and (7,0) the
major axis is 14 and half that is 7, so a = 7
Half the conjugate axis is half of 10, or 5.
So b = 5.
So the equation is:
or