Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Conic Section : Ellipse

Questions

Symbol Definition

Example : x^2 = square of x

Q01 | - Definition and diagrams

Q02 | - Equations of ellipse in rectangular coordinates

Q03 | - Equations of ellipse in polar form

Q04 | - Equations of ellipse in parametric equation

Q05 | - Example : Find equation for PF + PG = 10 if F and G are fixed

Q06 | - Example : Find coordinates of foci of ((x-2)/3)^2 + ((y+3)/5)^2 = 1

Q07 | - Example : Sketch ellipse

Q08 | - Example : Convert (x/5)^2 + (y/3)^2 = 1 to polar form

Q09 | - Example : convert R = 1.6/(1-0.8*cos(A)) to (x/a)^2 + (y/b)^2 = 1

Q10 | - Example : Change F(x ,y) = 0 to standard form

Q11 | - Formula of ellipse

Q12 | - Reference for ellipse

Q13 | - Prove the locus of ellipse is (x/a)^2 + (y/b)^2 = 1

Q14 | - Convert (x+f)^2/a^2 + y^2/b^2 = 1 to polar form

Q15 | - Draw tangent to ellipse by law of reflection

Q16 | - Prove that D*e = (a-f)*(1+e)

Q17 | - Elliminate x*y terms in F(x,y)

Q18 | - Compare polar form with standard rectangular form

Q19 | - Quiz and Answer

Q20 | -

Answers

Q1. Defintion of locus of an ellipse in rectangular system

Diagrams

Ellipse Locus
Ellipse Diagram of (x/a)^2 + (y/b)^2 = 1
Ellipse Diagram of (x/b)^2 + (y/a)^2 = 1
Ellipse Diagram of R = (D*e)/(1 - e*cos(A))
Ellipse Diagram of R = (D*e)/(1 + e*cos(A))

Defintion

Locus
- Two fixed points F and G which are the foci.
- A moving point P(x,y).
- When P moves so that PF + PG = constant = 2*a.
- The locus of p is an ellipse.
Equation of locus : ((x - h)/a)^2 + ((y - k)/b)^2 = 1.
- Center is C(h, k)
- Principal axis is y = k if a greater than b.
- The vertice on principal axis are U and V : CU = CV = a
- The foci on principal axis are F and G : focal length = CF = CG = f
Focal length
- f = Sqr(a^2 - b^2) where a greater than b.
- Eccentricity e = f/a

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Q2. Equations of ellipse in rectangular form

Standard Form

Equation : ((x - h)/a)^2 + ((y - k)/b)^2 = 1.
Where C(h, k) is the center. The values a and b are semi-axis.
Major axis and semi-axese
- If a is greater then b
  - then a is the major semi-axis.
  - then the principal axis is y = k.
  - then the foci are on principal axis.
- If a is less than b
  - then b is the major semi-axis.
  - then the principal axis is x = h.
  - then the foci are on principal axis.
The focal length is CF or CG which is f = Sqr(a^2 - b^2).
The vertex
- The end points of locus on principal axis are U and V. The a = CU = CV.
- The end points perpendicular to major axis is S and T. The b = CS = CT.
- FU = CU - CF = a - f.

Implicit Form without x*y

Equation : F(x,y) = A*x^2 + C*y^2 + D*x + E*y + F = 0. A and C have same sign.
Foci are on principal axis which is parallel to x-axis or y-axis.
Find foci, a, b, f of ellipse.
- Change to standard form by using completing the square.
- Then we get ((x - h)/a)^2 + ((y - k)/b)^2 = 1
- Hence we can find the center (h, k)
- Hence we can fond a and b
- Then we can find f

Implicit Form with term x*y

Equation : F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0.
B^2 - 4*A*C < 0 if it is an ellipse.
Principal axis is not parallel to x-axis or y-axis.
How to find semi-axis and foci ?
- We must elliminate x*y by rotating an angle.
- The new coefficients A' C' D' E' F' can be obtained with B' = 0.
- Then we can find a, b, f and coordinates of foci.
- Detailed method is given in ZE.txt or ZC.txt of MD2002.

Example : Study ((x + 2)/5)^2 + ((y - 3)/3)^2 = 1

Pricipal axis is y = 3.
Center at (-2, 3), a = 5 and b = 3.
Focal length f = Sqr(a^2 - b^2) = Sqr(5^2 - 3^2) = 4.
Foci are at (-6, 3) and (2, 3)
Vertice are at (-7, 3) and (3, 3)

Example : Study ((x + 2)/3)^2 + ((y - 3)/5)^2 = 1

Pricipal axis is x = -2.
Center at (-2, 3), a = 5 and b = 3.
Focal length f = Sqr(a^2 - b^2) = Sqr(5^2 - 3^2) = 4.
Foci are at (-2, 7) and (-2, - 1)
Vertice are at (-2, 8) and (-2, -2)

Example : Study 9*x^2 + 25*y^2 - 54*x + 100*y - 44 = 0

9*(x^2 - 6*x + 9) - 81 + 25*(y^2 + 4*y + 4) - 100 - 44 = 0.
9*(x - 3)^2 + 25*(y + 2)^2 = 225.
((x - 3)/5)^2 + ((y + 2)/3)^2 = 1.
Hence it is an ellipse.
- Pricipal axis is y = -2.
- Center at (3, -2), a = 5 and b = 3.
- Focal length f = Sqr(a^2 - b^2) = Sqr(5^2 - 3^2) = 4.
- Foci are at (-1, -2) and (7, -2)
- Vertice are at (-2, -2) and (8, -2)

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Q3. Equations in Polar Form

[Function 1]

Function : R = D*e/(1 - e*sin(A)).
- Origin is at F.
- Directrix is y = -D.
- D is focus to directrix line.
- Eccentricity e is less than 1.
- A is angle making with x-axis.
Foci are on y-axis and origin is on bottom focus.
Directrix is at bottom of ellipse.

[Function 2]

Function : R = D*e/(1 + e*sin(A)).
- Origin is at F.
- Directrix is y = D.
- D is focus to directrix line.
- Eccentricy e is less than 1.
- A is angle making with x-axis.
Foci are on y-axis and origin is on top focus.
Directrix is at top of ellipse.

[Function 3]

Function : R = D*e/(1 - e*cos(A)).
- Origin is at F.
- Directrix is x = -D.
- D is focus to directrix line.
- Eccentricy e is less than 1.
- A is angle making with x-axis.
Foci are on x-axis and origin is on left focus.
Directrix is at left of ellipse.

[Function 4]

Function : R = D*e/(1 + e*cos(A)).
- Origin is at F.
- Directrix is x = D.
- D is focus to directrix line.
- Eccentricy e is less than 1.
- A is angle making with x-axis.
Foci are on x-axis and origin is on right focus.
Directrix is at right of ellipse.

[Example] Relation of R = D*e/(1 - e*cos(A)) with ((x - f)/a)^2 + (y/b)^2 = 1

R = D*e/(1 - e*cos(A)) : Origin at F
((x-f)/a)^2 + (y/b)^2 = 1 and let it have same focus F.
When A = 180 degrees then cos(180) = -1.
Hence R = D*e/(1 + e).
From diagram we know R = FU = a - f when A = 180 degrees.
Hence D*e = (a - f)*(1 + e)
By defintion e = f/a

[Example] Prove that R = D*e/(1 - e*cos(A)) is ellipse

Construction
- Draw vertical line as directrix.
- Draw princial axis perpendicular to the directrix.
- Let F on princial axis as origin (0,0).
- Draw a point P(x,y).
- Let PQ = distance from P to Q where Q is on directrix.
- D = focus to directrix which is x = -D.
Locus of P is ellipse if PF/PQ = e where e is less than 1.
Prove that R = D*e/(1-e*cos(A)).
- By construction we know PF = R and PQ = D + x.
- D = distance from F to directrix.
- By defintion R/PQ = e.
- Hence R = e*PQ = e*(D + x).
- Since x = r*cos(A).
- Hence R = e*(D + e*R*cos(A))
- Hence R = e*D/(1 - e*cos(A))

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Q4. Equation of ellipse in parametric equation

x = h + a*cos(t).
y = k + b*sin(t).
(h,k) are center. The values of a and b are semi-axis.
Proof
- ((x - h)/a)^2 + ((y - k)/b)^2 = cos(t)^2 + sin(t)^2.
- Since cos(t)^2 + sin(t)^2 = 1.
- Hence ((x - h)/a)^2 + ((y - k)/b)^2 = 1.

Go to Begin

Q5. Two fixed point are F(-4,0) and G(4,0). A moving point is P(x,y). If PF + PG = 10, find the equation of the motion.

[Method 1] By defintion of standard equation

The motion is an ellipse and foci are on the x-axis.
The equation is ((x - h)/a)^2 + ((y - k)/b)^2 = 1.
- The center C is between F and G. Hence h = 0 and k = 0.
- Also f = CG = CF = 4.
- Since PF + PG = 2*a = 10 by defintion, Hence a = 5.
- Since focal f = Sqr(a^2 - b^2), hence b^2 = a^2 - f^2. and b = 3.
The requred equation is (x/5)^2 + (y/3)^2 = 1.

[Method 2] By using distance formula

Since PF + PG = 10.
Sqr((x + 4)^2 + y^2) + Sqr((x - 4)^2+ y^2) = 10
Sqr((x + 4)^2 + y^2) = 10 - Sqr((x - 4)^2+ y^2)
Square both sides we have
- (x + 4)^2 + y^2 = (10 - Sqr((x - 4)^2 + y^2))
- (x + 4)^2 + y^2 = 100 - 20*Sqr((x - 4)^2+ y^2)) + (x - 4)^2 + y^2
- (x + 4)^2 - (x - 4)^2 - 100 = -20*Sqr((x - 4)^2+ y^2))
- -16*x + 100 = 20*Sqr((x - 4)^2 + y^2))
- -4*x + 25 = 5*Sqr((x - 4)^2 + y^2))
Square both sides
- (-4*x + 25)^2 = 25*(x - 4)^2 + 25*Y^2
- 16*x^2 - 200*x + 625 = 25*x^2 - 200*x + 400 - 25*y^2
- -9*x^2 + 25*y^2 = -625 + 400
- 9*x^2 + 25*y^2 = 225
Simplify and we get
- (x/5)^2 + (y/3)^2 = 1.

Note

Method 1 is simple if we understand the defintion.
Method 2 is staightforward but the procedures are complicated.

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Q6. Example : If ((x-2)/3)^2 + ((y+3)/5)^2 = 1, find coordinates of foci

Solution

The foci are on principal axis which is x = 2.
The center is at C(2, -3)
Focal length f = Sqr(5^2 - 3^2) = 4
Coordinate of focus is at F(2, -7)
Coordinate of other focus is at G(2, 1)

Go to Begin

Q07. Example : Sketch an ellipse

[Method 1] Use a string Figure 3 Sketch by string

Let the string ends be fixed at F and G.
The length of the string is greater than FG.
Use a pencil as a point P on the string.
Let string be two sides of triangle PFG.
Move pencil and keep PFG as triangle.
The pencil will trace a triangle.

[Method 2] Use ruler and tractor Fifure 4 Sketch by ruler and tractor

Let F and G be two fixed points.
Draw a line FQ so that FQ = 2*a where a is the major semi-axis.
Join Q and G. Bisect QG and meet line FQ at P.
The P is a point on the ellipse.
Since bisector perpendicular to QG.
Hence GP = PQ and hence PF + PG = FQ = 2*a.
P is point on ellipse.
Repeat above step to find more points on ellipse.

[Method 3] Compute (x,y) using equation in rectangular form
[Method 4] Compute (x,y) using equation in parametric
[Method 4] Compute (R,A) using polar function
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Q8. Example : Convert x^2/5^2 + y^2/3^2 = 1 to polar form

Solution

The polar form is R = D*e/(1 - e*cos(A)) if left focus F is origin.
- When A=180 degrees, R = a - f.
- Hence a - f = D*e/(1 + e) or D*e = (a - f)*(1 + e)
x^2/5^2 + y^2/3^2 = 1 and left focus is at (-4, 0)
- Focal length f = Sqr(5^2 - 3^2) = 4.
- The eccentricity e = f/a = 4/5 = 0.8.
- D*e = (a - f)*(1 + e) = (5 - 4)*(1 + 0.8) = 1.8
Hence required polar function is R = 1.8/(1 - 0.8*cos(A))

Other method : See Q14.
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Q9. Convert R = 1.8/(1 - 0.8*cos(A) to (x/a)^2 + (y/b)^2 = 1

Polar form R = D*e/(1-e*cos(A)).
- D*e = 1.8 and e = 0.8.
- D*e = (a - f)*(1 + e) = 1.8,
- e = f/a = 0.8.
- Substitute f = 0.8*a into D*e.
- Hence (a-0.8*a)*(1+0.8) = 1.8.
- Hence 0.2*a = 1 and hence a = 5.
x^2/a^2 + y^2/b^2 = 1 : find b
- Since f = Sqr(a^2 - b^2) and hence b = 3.
- The required equation is x^2/5^2 + y^2/3^2 = 1.

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Q10. Example : 9*x^2 + 25*y^2 - 18*x - 100*y - 116 = 0, find a, b ,f

Change it to standard form by completing the square.
- 9*(x^2 - 2x + 1 -1) + 25*(y^2 - 4*y + 4 - 4) - 116 = 0.
- 9*(x-1)^2 - 9 + 25*(y-2)^2 - 100 - 116 = 0.
- 9*(x-1)^2 + 25*(y-2)^2 = 225.
Divide both sides by 225 and we have
Equation ((x - 1)/5)^2 + ((y - 2)/3)^2 = 1
Hence this is an ellispse
Hence a = 5 and b = 3. Then f = Sqr(a^2 - b^2) = 4.

Note

This is an ellipse because (B^2 - 4*A*C) = 0 - 4*9*25 = -900

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Q11. Formula :

(x-h)^2/a^2 + (y-k)^2/b^2 = 1.
- Principal axis is y = k if a is greater than b.
- Vertice on principal axis are U and V : CU = CV = a.
- Foci on principal axis are F and G : focal length = CF = CG = f.
- Focal length f = Sqr(a^2 - b^2).
- Efficiency e = f/a.
Locus in rectangular system
- F and G are foci.
- P is moving point.
- PF + PG = 2*a and locus of P is an ellipse.
Locus in polar system
- PQ is distance from P to directrix and Q on directirx.
- PF = R.
- R/PQ = e and locus of P is an ellipse if e is less than 1.
Relation between polar and rectangular system
- R = D*e/(1-e*cos(A))
- When A = 180 degrees and cos(A) = -1
- Hence R = (a - f) and R = D*e/(1 + e)
- Use polar formula we have D*e = (a - f)*(1 + e)
- Where a is majar semi-axis.
- D is the distance from focus F to directrix.
Slope of point on ellipse dy/dx = -(x*b^2)/(y*a^2).
Focal length f.
- CF = CG = f. C is center, F and G are foci.
- Eccentricity e = f/a.
- f = Sqr(a^2 - b^2).
- a - f = UF and U is vertex on principal axis near F
- D*e = (a - f)*(1 + e) when A = 180 in polar form.

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Q12. References :

Sketch (x-h)^2/a^2 + (y-h)^2/b^2 = 1 ..... MD2002 ZM40 02

Sketch R = D*e/(1-e*sin(A)) .............. MD2002 ZM40 08

Sketch F(x,y) = 0 ........................ MD2002

Conic Section ............................ MD2002 ZM06 and ZM40

Elliminate x*y terms in F(x,y)=0 ......... MD2002 ZM34 12

See keywords Ellipse in Index File
See contents of ZMxx in Contents by chapres

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Q13. Example : Prove that locus of ellipse is x^2/a^2 + y^2/b^2 = 1

Let the fixed points be F(x,-f) and G(x,f). Moving point is P(x,y).
Since PF + PG = 2*a where a is the major semi-axis.
Sqr((x+f)^2 + y^2) + Sqr((x-f)^2 + y^2) = 2*a.
Square bothe sides we have :
- (x+f)^2+ y^2+ (x-f)^2+ y^2+ 2*Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2)=4*a^2.
- x^2+2*x*f+f^2+y^2+x^2-2*x*f+f^2+f^2-4*a^2=-2*Aqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2).
- or 2*x^2+ 2*y^2+ 2*f^2- 4*a^2 = -2*Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2).
- or (x^2+ y^2)+ (f^2- 2*a^2) = -Sqr((x+f)^2+y^2)*Sqr(x-f)^2+y^2).
Square both sides again :
- (x^2+y^2)^2+2*(x^2+y^2)+(f^2-2*a^2)^2 = (x+f)^2+y^2)*(x-f)^2+y^2).
- x^4+2*x^2*y^2+y^4+ 2*f^2*(x^2+y^2)- 4*a^2*(x^2+y^2)+ f^4-4*f^2*a^2 +4*a^2.
- = (x+f)^2*(x-f)^2 + y^2*(x-f)^2 + y^2*(x+f) +y^4.
- = x^4-2*x^2*y^2+f^4+x^2*y^2-2*x*f*y^2+y^2*f^2+x^2*y^2+2*x*f*y^2+y^2*f^2+y^4.
Simplify above equation :
- 4*x^2*f^2 - 4*a^2*x^2 - 4*a^2*y^2 = 4*a^2*f^2 - 4*a^4.
- -(4*a^2-4*f^2)*x^2 - 4*a^2*y^2 = 4*a^2*(a^2-b^2) - 4*a^4 .
- -(a^2-f^2)*x^2 - a^2*y^2 = -a^2*b*2.
- b^2*x^2 + a^2*y^2 = a^2*b^2.
Hence equation of locus is x^2/a^2 + y^2/b^2 = 1.
By translation : (x-h)^2/a^2 + (y-k)^2/b^2 = 1.

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Q14. Example : Convert ((x + f)/a)^2 + (y/b)^2 = 1 to polar form

The focal point is at F(x,-f) for polar form R = D*e/(1-e*cos(A)).
Remove denominator :
- b^2*(x+f)^2 + a^2*y^2 = a^2*b*2.
- b^2*x^2 + b^2*x*f + b^2*f^2 + a^2*y^2 = a^2*b^2.
- Since b^2 = a^2 - f^2.
- Hence b^2*x^2 + a^2*y^2 + 2*b^2*x*f = a^2*b^2 - b^2*f^2.
- (a^2-f^2)*x^2 + a^2*y*2 + 2*b^2*x*f = b^2*(a^2-f^2)
Since x = R*cos(A) and y = R*sin(A) and R^2 = x^2 + y^2.
Simplify above equation :
- R^2*a^2 - f^2*x^2 + 2*b^2*x*f = b^4.
- R^2*a^2 - (f^2*x^2 - 2*b^2*x*f + b^4) + b^4 = b^4.
- R^2*a^2 - (f*x - b^2)^2 = 0.
- R*a = f*x - b^2.
- R*a = -(f*x - b^2).
- R*a - R*f*cos(A) = b^2
- R*(a - f*cos(A)) = b^2.
After elliminting f by f=e/a we have : R = b^2/(a - f*cos(A)).
Since f/a = e, hence R = (b^2/a)/(1 - e*cos(A)).
If A = 180 we have R = (b^2/a)/(1+e) or b^2/a = R*(1+e).
b^2/a = (a^2 - f^2)/a = (a-f)*(a+f)/a = (a-f)*(1+e) = D*e.
Hence R = D*e/(1-e*cos(A)).