drawing ellipses · 2019-02-05 · ellipse is translated so that x=x!+h and y=y!+k, where (x!,y!)...

ARML Power Contest – February 2010 =========================================================

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Drawing Ellipses The Background The study of conic sections, curves formed by the intersection of a plane with a cone, began about 200 B.C. by Apollonius of Perga, the last of the great mathematicians of the Golden Age of Greek mathematics. In his eight-volume treatise called “Conics,” he gave names to the various curves (parabola, hyperbola, and ellipse) and developed most of the geometric properties of the curves. But conics remained pure mathematics for almost 2000 years until 1609 when Johannes Kepler revolutionized astronomy by declaring that the planets revolved around the sun in elliptic paths with the sun as a focus. As an ellipse is a circle stretched horizontally and/or vertically, ellipses share some similar formulas with circles. For example, the area of a circle is ! " r " r while the area of an ellipse is ! "a "b . However, while the circumference of a circle is 2 !" ! r = " (r + r) , the circumference of an ellipse is not ! (a + b) and cannot even be expressed in a closed form. I find it amazing that both a cone and a cylinder cut diagonally by a plane produce an ellipse, and that if you wrap a strip of paper around a cylinder, slice the cylinder diagonally, and then unwrap the paper, the edge of the paper formed by the circumference of the ellipse is a sine wave! There is much to explore in this old topic of mathematics! Definitions for an ellipse: 1. The curve formed when a plane intersects a right circular cone at an angle to its axis of symmetry less than 90° but greater than the slant angle of the cone. 2. The set of all points in a plane the sum of whose distances from two given points (called the foci) is always constant. 3. The set of all points whose ratio of its distance from a fixed point (called the focus) to its distance from a fixed line (called the directrix) is constant and between 0 and 1. 4. The graph formed by the general second-degree equation Ax2 + Bxy+Cy2 + Dx + Ey+ F = 0 where 4AC ! B

2> 0 .

5. The curve formed when a cylinder is intersected by a plane not parallel to the axis of symmetry of the cylinder. Other vocabulary: A segment connecting two points of an ellipse is called a chord. A chord through the center of an ellipse is called a diameter. The longest diameter of an ellipse is called the major axis. The foci will always be on the major axis. The shortest diameter of an ellipse is called the minor axis. It will always be perpendicular to the major axis. The focal length is the distance between the two foci. The eccentricity of an ellipse, a number between 0 and 1, refers to the “roundness” of the ellipse. The closer the eccentricity is to 0, the closer the ellipse is to a circle.


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Equations of an ellipse: 1. General equation: Ax2 + Bxy+Cy2 + Dx + Ey+ F = 0 where 4AC ! B

2> 0 . If the

ellipse is centered at the origin, then D = E = 0. If the major and minor axes of the ellipse are parallel with the coordinate axes, then B = 0.

2. Standard form: x2

a2+y2

b2=1 . The length of the major axis is 2a and the length of the

minor axis is 2b . The focal length is 2c , where c2 = a2! b

2 . The eccentricity e is ca

. If the

ellipse is translated so that x = !x + h and y = !y + k , where ( !x , !y ) is the point on the translated

coordinate axes, it is now centered at (h,k) and the equation becomes (x ! h)2

a2+(y! k)2

b2=1 . If

the ellipse is rotated an angle of ! , then x = !x cos" # !y sin" and y = !x sin" + !y cos" and the general equation must now be used to describe the curve.

3. Parametric equations: x = acos(!)y = bsin(!)

"#$

%&'

.

4. Polar equation: r = ep

1! ecos" with directrix x = ± p or r = ep

1! esin" with directrix

y = ± p . One focus is at the pole and 0 < e < 1 is the eccentricity of the ellipse. Through various algebraic techniques these equations can be shown to be equivalent. The problems in this contest deal with methods that are used by artists, engineers, draftsmen, astronomers, and mathematicians for drawing ellipses. In several problems you are given the method of construction and must prove that the curve formed is an ellipse. This can be done by showing that points on the curve satisfy one of the above equations or that the curve has one of the properties that defines an ellipse. Most of the problems are independent of one another and therefore can be solved in any order.


3

The Problems 1. An ellipse can easily be drawn using two push-pins and a loop of string as shown at the right. By keeping the triangular loop of the string taut as you move the pencil, a smooth curve can be drawn.

a) How do you know the curve formed is an ellipse? b) If you want to draw an ellipse with a major axis of length 12 and a minor axis of length 8, how far apart should the push-pins be placed and what should be the length (or circumference) of the loop of string?

2. An ellipse can made by drawing a large circle on a piece of wax paper and marking a point F somewhere inside the circle but not at the center. Fold the paper so that some point P on the circle coincides with F and crease the paper on this fold. Unfold the paper and repeat the process using another point P on the circle. Continue repeating the process.

a) Using the diagram at the right, prove that the points Q determined by this technique trace out an ellipse. b) If circle O, centered at the origin, has a radius of 10 units and F is 8 units from the center, determine an equation for the ellipse formed by this technique.

OF

P1

Q1

l1


4

3. On a set of coordinate axes, draw a circle C1 centered at the origin and having a radius equal

to a, and a second circle C2 centered at the origin and having a radius equal to b, where a > b .

Draw a line through the origin (at an angle of ! ), intersecting C1 and C

2 at points M and N,

respectively. Draw a line parallel to the vertical axis through M and another line parallel to the horizontal axis through N. Label the intersection of these two lines P. Repeat the process above for different values of ! , determining a locus of points P. Connect all the points P to form an ellipse.

a) If P = (x, y) , show that it lies on the curve

defined by x2

a2+y2

b2=1 .

b) If the area between the two circles (called an annulus) is equal to the area of the ellipse,

what is the ratio of a

b?

4. The Dutch mathematician Franz Van Schooten came up with the following simple device for drawing ellipses. Points A and B are stationary, with AB > 4 !CD . CD = DE and points C and D are hinges with C anchored at the midpoint between A and B and D is free to move as point E slides in the groove between A and B. At point P is a pen which traces out a curve as E moves from A to B.

a) Let CD = a and DP = b. If C is the origin and if this curve is an ellipse, what would be its equation? b) Let P = (x, y) and E = (t,0) . Prove P(x, y) satisfies the equation above for any value of t in the interval [-2a, 2a].

N

M

P

!

C2

C1

P

E

D

C

BA


5

5. Using a Spirograph, a circular disk can be rolled inside of a fixed circular ring without slipping, producing a figure called a trochoid. A point P on the disk marks the position of a pen which traces out a curve as the disk is rolled. If the inner radius of the fixed circular ring is R, the radius of the circular disk is r, and d is the distance from P to the center of the disk, the

coordinates of P are (x, y) , where x = (R ! r)sinr

R"#

$%&'(+ d sin

r ! RR

#$%

&'("

#$%

&'(

and

y = (R ! r)cosr

R"#

$%&'(+ d cos

r ! RR

#$%

&'("

#$%

&'(

. (For a nice derivation of these formulas see

html:/the-robinson-family.org/Nigel/spiro.htm.) The figure below shows four trochoids and the ratio r:R that produced them.

If R =2r, prove the trochoid is an ellipse.


6

X

Y Z

6. Let O be the origin, A = (a,0) , B = (0,b) , C = (0,!b) , and D = (a,b) . Divide segments OA and AD into n congruent segments using n – 1 points. On AD , starting at A, label these points D

1, D

2, D

3, ... , D

k, ... , D

n!1. On OA ,

starting at A, label the points A

1, A

2, A

3, ... , A

k, ... , A

n!1. Let (x, y) be

coordinates of point Pk, the intersection of

lines CA

k

! "###and

BD

k

! "###.

Prove Pk is on ellipse x

2

a2+y2

b2=1 . (Hint:

One method might be to find the equations of lines

CA

k

! "### and

BD

k

! "### by first expressing

the coordinates of Dk and A

k in terms of a,

b, n, and k.) 7. Take any triangle XYZ and position it on a coordinate axes so that X is on the x-axis and Y is on the y-axis. Mark the location of point Z. Reposition the triangle so that again X is on the x-axis and Y is on the y-axis and plot point Z. Repeat several times. The locus of all such points Z will be an ellipse, whose axes are not parallel with the coordinate axes.

a) If XYZ is an equilateral triangle with sides of length 2, what is the equation of the ellipse formed? (Hint: Determine the exact coordinates of some points on the ellipse and determine the general equation.) b) If XYZ is a 30-60-90° triangle with angle Z = 30° and hypotenuse XZ of length 2, what is the equation of the ellipse formed?

O A

B

C

D

P1

P2

P3

D3

A3

D1

A1

A2

D2

Construction with

n = 4


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8. Given a line D (called the directrix) and a point F (called the focus). Let P be any point (x, y). If PF is the distance from the point to the focus and PD is the distance from the point to the directrix, then an ellipse is the set of all points P where the ratio of PF to PD is a constant less than 1. The ratio PF/PD is called the eccentricity of the ellipse.

a) Use the graph below to accurately plot at least 10 points of the ellipse with an eccentricity of 2

3. (Hint: 2

3=

4

6=

3

4.5= ... ) (Use Special Answer Sheet #1 to record your

answers to 8a and 8b.) b) If F is the origin and D is the line y = !3 , what is the equation of this ellipse?

10

9

8

7

6

5

4

3

2

1

0

1

F

D

1 2 3 4 5 6


8

9. Logan Graphics Products Inc. makes an Oval Mat Cutter used in picture framing. I was curious whether the oval it produced was an ellipse or not. Pictures of the tool did not reveal how it worked but designer Curt Logan was gracious enough to send me a free mat cutter to examine for myself. Although the cutting head is a patented part, inside the oval base I discovered that its mechanism was similar to a trammel for drawing ellipses designed by Archimedes! As point P is rotated around the origin, point Q moves back and forth in a horizontal (x-axis) channel while point R moves up and down in a vertical (y-axis) channel. If RQ = a, QP = b, Q = (t,0) , and P = (x, y) , show that the locus of all possible points P determine an ellipse by determining the equation of the ellipse.

Type set using MathType 5.0, donated by Design Science, an ARML supporter.

P

Q

R


9

The Solutions 1a. Let L = the length of the loop of string.

L = F1P + F

2P + F

1F2

L ! F1F2= F

1P + F

2P

constant = F1P + F

2P

Therefore, P is on an ellipse. 1b.

2a. For any point P

1 on the circle, P

1 is folded onto F forming l

1 , the perpendicular bisector of

segment FP1. Therefore, Q

1F = Q

1P . Since OP

1 is the radius of the circle and

OP1= OQ

1+Q

1P1, OQ

1+Q

1P1 is constant and therefore, OQ

1+Q

1F1 is constant. Therefore, Q is

a point on an ellipse with foci at O and F1.

b. 2a = 10 and 2c = 8 . Therefore, a = 5 and c = 4 and since a

2! b

2= c

2, b = 3. Assume F is

(8,0), then the center is at (4,0). Therefore, the equation of the ellipse

is x ! 4( )

2

25+y2

9= 1 .

3a. cos(!) =x

a" cos

2(!) =

x2

a2

and sin(!) =y

b" sin

2(!) =

y2

b2

.

Adding these two equations produces 1 =x2

a2+y2

b2

.

3b. The area of the ellipse = ab! and the area of the annulus = a2! " b

2! . Therefore,

ab = a2 ! b2

ab

b2=a

2

b2!b

2

b2

a

b=a

2

b2!1

0 =a

b

"#$

%&'

2

!a

b

"#$

%&'!1

a

b=

1+ 5

2, the Golden Ratio!

L = F1P + F

2P + F

1F2

= 2a + 2c

= 12 + 2c

a2! b

2= c

2

36 !16 = c2

2 5 = c

F1F2= 2c = 4 5 ! 8.94

L = 12 + 4 5 ! 20.94

θ


10

4a. When CD is horizontal CP = a + b and when CD is vertical CP = a ! b . Therefore,

x2

a + b( )2+

y2

a ! b( )2= 1 .

4b. From similar right triangles,

b

x !t

2

=a ! b

t ! x" t =

2ax

a + band from the Pythagorean

Theorem, t ! x( )2

+ y2= a ! b( )

2 . Substituting, results

in 2ax

a + b! x"

#$%&'2

+ y2= a ! b( )

2 . And so

2ax

a + b!x(a + b)

a + b

"#$

%&'2

+ y2= a ! b( )

2 (x(a ! b)a + b

"#$

%&'2

+ y2= a ! b( )

2 ( . Resulting in

x2(a ! b)

2

a + b( )2

+ y2= (a ! b)

2"

x2

a + b( )2+

y2

(a ! b)2= 1 .

5. If R = 2r then x = r sin!2

"#$

%&'+ d sin

(!2

"#$

%&'

and y = r cos!2

"#$

%&'+ d cos

(!2

"#$

%&'

. This simplifies to

x = r ! d( )sin"2

#$%

&'(

and y = r + d( )cos!2

"#$

%&'

. Squaring, results in x2 = r ! d( )2

sin2

"2

#$%

&'(

and y2 = r + d( )2

cos2

!2

"#$

%&'

, and dividing, produces x2

r ! d( )2= sin

2"2

#$%

&'(

and

y2

r + d( )2= cos

2!2

"#$

%&'

. Summing, results in x2

r ! d( )2+

y2

r + d( )2= 1 .

6. First, find the coordinates of Ak and D

k.

Ak= 1!

k

n

"#$

%&'a, 0

"#$

%&'

and Dk= a,

k

n

!"#

$%&b

!"#

$%&

.

Second, find the equations of CA

k

! "### and

BD

k

! "###.

CAk

! "###:nbx + (k ! n)ay = ab(n ! k) and

BDk

! "###: n ! k( )bx + any = abn .

Square each: n2b2x2 + 2nbx(k ! n)ay + (k ! n)2a2y2 = a2b2 (n ! k)2 and n ! k( )

2b2x2+ 2(n ! k)bxany + a

2n2y2= a

2b2n2 .

Add them, producing: n2 + n ! k( )2( )b2x2 + n

2+ (k ! n)

2( )a2y2 = a2b2 (n ! k)2 + n2( ) .

Divide by a2b2 (n ! k)2 + n2( ) , resulting in: x2

a2+y2

b2= 1 (N.B. (n ! k)2 = (k ! n)2 .)

C

D

E

a

t

x

y a-b

b

t-x

x !t

2

P=(x,y)


11

7. Let the equation of the ellipse be: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 . Since it is centered at the origin, D = E = 0 . To avoid answers that are just multiples of each other, let A = 1. Three points are necessary to solve the equation x2 + Bxy + Cy2 + F = 0 . Positioning !XYZ as in the diagrams produces these three points:

Solving this system of equations,

1+ B 3( ) + C(3) + F = 0

3+ B 3( ) + C(1) + F = 0

4 + B 2 3( ) + C(3) + F = 0

!

"

##

$

##

%

&

##

'

##

, results in

x2! 3xy + y

2!1 = 0 as the equation of the ellipse.

7b. As in solution 7a, three points are needed to satisfy the equation x2 + Bxy + Cy2 + F = 0 .

Positioning !XYZ as in the diagrams produces these three points:

Solving this system of equations,

0 + B 0( ) + C(3) + F = 0

3+ B 3( ) + C(1) + F = 0

9

4+ B 0( ) + C(0) + F = 0

!

"#

$#

%

&#

'#

, results in

x2!

3

2xy +

3

4y2!9

4= 0 or 4x2 ! 2 3xy + 3y

2! 9 = 0 as the equation of the ellipse.

8a. (Drawing follows solution to problem #9.) 8b. The length of the major axis is 6 + 1.2 = 7.2 and so b = 3.6, the center is (0, 2.4), and c = 2.4.

Since a2 = b2 ! c2 , a2 = 7.2. Therefore, the equation of the ellipse

isx2

7.2+

y ! 2.4( )2

12.96= 1 .

X

Z= 0, 3( )

Y

Z= 3,1( )Y

XZ= 1.5, 0( )

Y

X

Z = 1, 3( )Z = 3,1( )

Z = 2, 3( )


12

9. From the diagram below, x ! t( )2

+ yy= b

2 and a

t=

b

x ! t" t =

ax

a + b. Substituting,

x !ax

a + b

"#$

%&'2

+ y2= b

2

!x(a + b) " ax

a + b

#$%

&'(2

+ y2= b

2

!bx

a + b

"#$

%&'2

+ y2= b

2

!b2x2

a + b( )2+ y

2= b

2

!x2

a + b( )2+y2

b2= 1

8a.

10

9

8

7

6

5

4

3

2

1

0

1

F

D

1 2 3 4 5 6

P=(x,y)

Q=(t,0)

a

y

x - t

t

O=(0,0)

b

drawing ellipses · 2019-02-05 · ellipse is translated so that x=x!+h and y=y!+k, where (x!,y!)...

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