Algebra II
Chapter 10 Section 5Hyperbola
Hyperbola
The set of all points that have the difference of the distance to 2 focus points is a given constant
Parts of a hyperbola
Transverse Axis
Foci
Asymptotes
Vertices
Standard Form of an Hyperbola
If a & b > 0 the Transverse axis is Horizontal
If a & b < 0 the Transverse axis is Vertical
12
2
2
2
b
y
a
x
Put into standard Form
3649 22 yx
Put into standard Form
324936 22 yx
In standard form• The +-value for a (square root of the
denominator of x2) are the vertex’s x if the Transversal Axis is horizontal, values (-a,0) and (a,0)
• The +-value for b (square root of the denominator of y2) are the vertex’s y if the Transversal Axis is Vertical, values (0,-b) and (0,b)
To find the Asymptote
xa
by
xa
by
Find the Asymptotes and Vertices
12516
22
yx
1949
22
xy
12516
22
yx
xa
by
Horizontal Transverse Axis
So Vertices are at +- (a,0)
a = 4 b = 5
(4,0) and (-4,0)
Asymptotes are at
xy4
5 xy
4
5
1949
22
xy
Given a vertex at (-3, 0) and an asymptote of y = 4x
Find the equation of hyperbola
Foci
• Foci are always on the transverse axis
• Foci are equidistant from the minor axis
• The distances to a focus from the origin is equal to the distance along the asymptote, with a coordinate = to the vertex’s non zero coordinate
Find the focus of an Hyperbola with its center on the origin
• If c is the distance from the center to the focus
• Find c using c2 = a2 + b2
• if your transverse axis is Horizontal your foci are at (+-c ,0)
• if your transverse axis is Vertical your foci are at (0 ,+-c)
a & b are + so we have a horizontal Transverse axis (y = 0)
c2 = a2 + b2
c2 = 16 + 9c2 = 25c = 5
So the coordinates of the foci are at (5,0) and (-5,0)
Find the foci of 1
916
22
yx
125144
22
xy
Find the Foci
Find the equation given a vertex and a focus
• Find the other vertex using c2 = a2 + b2
• Substitute in the 2 values you know and solve for the 3rd
– Remember the foci will lie along the Transverse axis
– c is the distance of the focus to the origin
• Once you have the 2 vertices, take the square roots to get a and b & substitute them in
Given a focus of (10,0) and a vertex at (-6,0) find the equation of the
Hyperbola
The transverse axis is Horizontal since the vertex and focus lie on the x axis
Since the vertex is at (0,-6) a2 = 36Since the focus is at (0,10) c2 = 100
Substitute a2 and c2 into a2 + b2 = c2 and solve for b36 + b2 = 100 b2 = 64 b= 8
Substitute a2 & b2 into 12
2
2
2
b
y
a
x1
6436
22
yxTo
get
Given a focus of (13,0) and a vertex at (5,0) find the equation of the ellipse
Given a focus of (0,4) and a vertex at (0,5) find the equation of the ellipse
Do Now!
• Page 566 Problems 2 – 18 even, find the vertex, foci and
asymptotes as well as sketch a graph
23 – 26 all30, 31, 36, 40, 42, 49-51