Download Section 23.1

1. Show geometrically that given any two points in the hyperbolic
plane there is always a unique line passing through them.
§ 23.1
Given point A(x1, y1) and B(x2, y2). These points can always be
substituted into the general half-circle with center on the x-axis.
x 2 + y 2 + ax = b
The resulting two equations (one from each point) can be solved except in
the case where x1 = x2. However, in this case, the line is a vertical ray
with the equation x = x1.
2. Given two points A and B on the hyperbolic with point C between them that the
betweenness property holds. I.e. verify that if A – B – C then AC + CB = AB.
Note there are two cases for the two “different” types of lines in this model
Case 1 – points on a ray: A (a, m), B (b, m) and C (c, m)
a
c
AC  CB  ln    ln  
c
b
 a  c 
a
 ln      ln    AB
 c  b 
b
Case 2 – points on a semicircle:
 AM CN 
 CM BN 
AC  CB  ln 
  ln 

 AN CM 
 CN BM 


 AM CN CM BN 
ln 

 AN CM CN BM 
 AM BN 
ln 
  ln (AB, MN)  AB
 AN BM 
3. Let A = (1, 3) B = (1, 6), and C = (5, 7)
a. Find the equation of the line AB.
b. Find the equation of the line AC
a. X coordinates are the same. It is a vertical line with equation x = 1.
b. Substitute (1, 3) and (5, 7) into x2 + y2 + ax = b and solve the resulting two
equations
10 + a = b and 74 + 5a = b
for a and b yielding a = - 16 and b = - 6. hence the equation of the line AC is
x2 + y 2 – 16 x = - 6
4. Let A = (3, 1), B = (3, 10), C = (12, 20) and D = (24, 16)
a. Find the distance from A to B.
b. Find the distance from C to D.
a. It is a ray so distance AB = 2.30
 10 
AB  ln    2.30
 1
b.
Not a ray so distance CD = 0.69
You need to find M and N first. Use the method of problem 3 to find the
equation of the line CD. x 2 + y 2 - 24x = 256. this circle has x-intercepts at
M = 32 and N = - 8.
 CM DN 
CD  ln 
 
CN
DM



 202  202
ln 
 202  202

 (12  32)2  202 (24  (  8))2  162 

ln 
 (12  (  8))2  202 (24  32)2  162 


322  162 
 800 1280 
1
 
ln 

2
82  162 
 800 320 

1
ln 4  0.69
2
5. Find the angle between the two h-lines x 2 + y 2 = 25 and x 2 + y 2 – 10x = - 9
Solve the two equations simultaneously to get the point of intersection at I(3.4, 3.6661).
I like to use the derivative of the lines to get the slopes at the point of intersection.
First Line: dy/dx = - x/y = - 0.9274 = m 1
Second Line : dy/dx = (5 – x)/y = 0.4364 = m 2
Tan  = (m 2 – m 1)/(1 + m 1 m 2) = 1.3638/.5953 = 2.2910
And  = 66.4218
You may also make an accurate drawing and use a protractor to measure the
angle.
6. Show that the two lines x = 5 and x 2 + y 2 – 6x + 5 = 0 are parallel.
a. Substituting x = 5 into the second equations yields a y value of 0.
this means that both lines have the point (5, 0) in common. This point
is on the x-axis and is an ideal point (at infinity) and the two lines are
parallel.
7. Let A = (2, 1), B = (2, 3), C = (2, 14) and D = (2, 16). Segments AB and CD
have the same Euclidean length. Find the hyperbolic lengths of the two
segments and compare.
AB = 1.10 and CD = 0.13. What is going on with distance in this
“hyperbolic” model?
8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and
C = (1, 10).
a. Calculate AC.
b. Calculate BC
c. Verify that ABC is an isosceles right triangle.
d. Verify that the point D(-24, 0) is the center of the semicircle AB, and
use this to calculate each of the angles of ABC using the slope formula for tan .
What is the angle sum of the triangle?
Consider making a drawing of this triangle.
a. AC = 0.6931
b. BC = 0.6931
c. See parts a and b. I will confirm it is a right triangle in part d.
d. The equation for the three sides are:
AB: x 2 + y 2 + 48x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1
Note that the center AB is (- 24, 0). It is always (-a/2, 0)
The center of AC is (1, 0) and thus AC and BC form a right angle at vertex
C.
continued
8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and
C = (1, 10).
d. Verify that the point D(-24, 0) is the center of the semicircle AB, and
use this to calculate each of the angles of ABC using the slope formula for tan .
What is the angle sum of the triangle?
d.
The equation for the three sides are:
AB: x 2 + y 2 + 48x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1
 C = 90. I will use the method of problem 5 to find the other two angles.
To find  A use sides AB and AC.
First Line, AB: dy/dx = x + 24/- y = - 3.875 = m 1 at point A
Second Line, AC : dy/dx = (1 – x)/y = - 0.75 = m 2 at point A
Tan  = (m 2 – m 1)/(1 + m 1 m 2) = - 3.125/3.90625 = - 0.8
And  =  A = 38.6598
continued
8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and
C = (1, 10).
d. Verify that the point D(-24, 0) is the center of the semicircle AB, and
use this to calculate each of the angles of ABC using the slope formula for tan .
What is the angle sum of the triangle?
d.
The equation for the three sides are:
AB: x 2 + y 2 + 48x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1
 C = 90 and  A = 38.6598.
To find  B use sides AB and BC.
First Line, AB: dy/dx = x + 24/- y = - 1.25 = m 1 at point B
Second Line, BC: BC is a vertical line with slope undefined. To find  B we will
first find the angle between line AB and a horizontal line (m 2 = 0) at point B.  B
will be 90 – the angle we find.
Tan  = (m 2 – m 1)/(1 + m 1 m 2) = 1.25
And  = 51.3402
 B = 90 – 51.3402 = 38.6598.
8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and
C = (1, 10).
d. Verify that the point D(-24, 0) is the center of the semicircle AB, and
use this to calculate each of the angles of ABC using the slope formula for tan .
What is the angle sum of the triangle?
d.
The equation for the three sides are:
AB: x 2 + y 2 + 48x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1
 C = 90 and  A = 38.6598. and  B = 38.6598.
Notice that the angles opposite equal sides are equal.
The sum of the angles of ABC is 167.3196.
9. Using the information from the previous problem does the Pythagorean Theorem
hold in Hyperbolic Geometry?
BC = a = 0.6931, AC = b = 0.6931, and AB = c = 1.0163
a 2 + b 2 = 0.9608 while c 2 = 1.0329. The Pythagorean Theorem does not
hold in hyperbolic geometry.
10. Using the information from the previous problem does the following relationship
cosh c = cosh a cosh b hold?
BC = a = 0.6931, AC = b = 0.6931, and AB = c = 1.0163
cosh a cosh b = 1.5625 and cosh c = 1.5624 Other than round off error the
relationship holds. Strange “Pythagorean Theorem”!!!!
11. Find the two h-lines parallel to x 2 + y 2 = 25 through the point (5, 10).
Consider making a drawing of this triangle.
The line x 2 + y 2 = 25 has center at the origin and radius 5. thus it crosses the xaxis at (-5, 0) and (5, 0). Each of lines we need will go through one of these
points and the given point (5, 10).
First line. Through (5, 10) and (-5, 0). Using the method of problem 3
substituting into x2 + y2 + ax = b gives
125 + 5a = b and 25 – 5a = b which solve giving a = - 10 and b = 75
So the lines is x2 + y2 - 10x = 75
Second line. Through (5, 10) and (5, 0). Notice that the x values of these two
points are the same so it is a vertical ray with equation x = 5.