1. 10-5 Hyperbolas
Hubarth
Algebra II

2. hyperbola. The midpoint of the segment is the center of the hyperbola.
A hyperbola is a set of points 𝑃 in a plane such that the absolute value of the difference
between the distances from 𝑃 to two fixed points 𝐹 1 and 𝐹 2 is a constant 𝑘.
𝑃 𝐹 1 −𝑃 𝐹 2 =𝑘, where 𝑘< 𝐹 1 𝐹 2
Each fixed point F is a focus of a hyperbola. The segment that lies on the line containing the foci
and the endpoints on a hyperbola is the transverse axis. The endpoints are the vertices of a
hyperbola. The midpoint of the segment is the center of the hyperbola.
𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1
Standard form of an equation of a hyperbola with a horizontal transverse axis

3. 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1
Standard form of an equation of a hyperbola with a horizontal transverse axis
𝒚= 𝒃 𝒂 𝒙
𝒚=− 𝒃 𝒂 𝒙
foci: 𝐹 1 , 𝐹 2
vertices: ±𝑎, 0
asymptotes: 𝑦=± 𝑏 𝑎 𝑥
x-intercept: ±𝑎
y-intercept: none . .
(-a, 0)
(a, 0)
𝑭 𝟏
𝑭 𝟐
Transverse Axis
𝑦 2 𝑎 2 − 𝑥 2 𝑏 2 =1
Standard form of an equation of a hyperbola with a vertical transverse axis
foci: 𝐹 1 , 𝐹 2
vertices: (0, ±𝑎)
asymptotes: 𝑦=± 𝑏 𝑎 𝑥
x-intercept: none
y-intercept: ±𝑎 .
𝑭 𝟏
𝒚=− 𝒃 𝒂 𝒙
𝒚= 𝒃 𝒂 𝒙
(0, a) .
(0, -a)
𝑭 𝟐
Transverse Axis

4. Ex. 1 Graphing Hyperbolas
Graph 4x2 – 16y2 = 64.
The equation of the form 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1,
so the transverse axis is horizontal.
4x2 – 16y2 = 64
– = 1
x 2 16
y 2 4
Since a2 = 16 and b2 = 4, a = 4 and b = 2.
Step 1: Graph the vertices. Since
the transverse axis is horizontal, the
vertices lie on the x-axis. The
coordinates are (±a, 0), or (±4, 0).
Step 2: Use the values a and b to draw the
central rectangle. The lengths of its sides
are 2a and 2b, or 8 and 4.
Step 3: Draw the asymptotes. The
equations of the asymptotes are
𝑦=± 𝑏 𝑎 or 𝑦=± 1 2 𝑥. The asymptotes
contain the diagonals of the central
rectangles
Step 4: Sketch the branches of the hyperbola
through the vertices so they approach the
asymptotes.
Quick Check

5. Ex. 2 Finding the Foci of a Hyperbola
Find the foci of the graph 𝑦 2 4 − 𝑥 2 9 =1. Draw the graph
The equation is in the form – = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
= Substitute 9 for a2 and 4 for b2.
c = Find the square root of each side of the equation.
The foci (0, ±c) are approximately (0, –3.6) and (0, 3.6). The vertices (0, ±b) are (0, –2) and (0, 2).
The asymptotes are the lines y = ± x , or y = ± x. b a 2 3

6. Ex. 3 Real-World Connection
As a spacecraft approaches a planet, the gravitational pull of the planet changes the spacecraft’s path to a hyperbola that diverges from its asymptote. Find an equation that models the path of the spacecraft around the planet given that a = 300,765 km and
c = 424,650 km.
Assume that the center of the hyperbola is at the origin and that the transverse axis is horizontal. The equation will be in the form – = 1.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
(424,650)2 = (300,765)2 + b2 Substitute.
1.803  1011 =  b2 Use a calculator.
b2 =  1011 –  1010 Solve for b2.
=  1010
The path of the spacecraft around the planet can be modeled by
𝑥 ∗ − 𝑦 ∗ =1

7. 1. Graph the hyperbola with equation 𝑦 2 16 − 𝑥 2 9 =1
Practice
1. Graph the hyperbola with equation 𝑦 − 𝑥 2 9 =1
2. Find the foci of 𝑥 − 𝑦 2 9 =1. Draw the graph.
34 , 0 , (− 34 , 0)
3. Find an equation that models the path of Voyager 2 around Jupiter, given that
a = 2, 184,140 km and c = 2, 904, km
𝑥 ∗ − 𝑦 ∗ =1