Section 23.1
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Transcript 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.