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
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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
Go to Begin
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)
Go to Begin
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))
Go to Begin
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
Note
- Method 1 is simple if we understand the defintion.
- Method 2 is staightforward but the procedures are complicated.
Go to Begin
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
Go to Begin
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.
Go to Begin
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.
Go to Begin
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
Go to Begin
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.
Go to Begin
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
Go to Begin
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.
Go to Begin
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)).
Other method : See Q08
Go to Begin
Q15. Example : Draw tangent to ellipse using reflection
Ellipse
Tangent by reflection
- Draw an ellipse with F(x,-f) and G(x,f).
- Draw a point P(x,y) on the ellipse.
- Draw a bisector of angle FPG.
- Draw a line perpendicular the bisector and passing P.
- By the law of reflection, This line is the requred tangent.
Go to Begin
Q16. Example : Relation of D*e = (a-f)*(1+e) in polar form
- Draw an ellipse with F(x,-f) and G(x,f) on x-axis.
- Let the vertex U be between directrix and F.
- When A = 180 then R = D*e/(1+e) = FU = a-f.
- Hence D*e = (a-f)*(1+e).
- Where e = f/a and f = Sqr(a^2-b^2).
Go to Begin
Q17. Elliminate x*y terms in F(x,y)
- Equation : A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0.
- Use rotation formula to change above equation as
- Equation : A'*x^2 + B'*x*y + C'*y^2 + D'*x + E'*y + F' = 0.
- Let B' = 0 and then we can find the rotation angle.
- Then we can find A', C', D', E', F'.
- Then we can find a, b, f, and the cneter (h,k) and principal axis.
-
Examples
in Conic Sections
Go to Begin
Q18. Compare polar form with rectangular standard form
R=D*e/(1-e*cos(A))
- Directrix at left side. Principal axis is x-axis. Focus F is origion.
- F is left side focus of x^2/a^2 + y^2/b^2 = 1.
- When A=180 we have D*e = (a-f)*(1+e)
R=D*e/(1+e*cos(A))
- Directrix at right side. Principal axis is x-axis. Focus G is origion.
- G is right side focus of x^2/a^2 + y^2/b^2 = 1.
- When A=0 we have D*e = (a-f)*(1+e)
R=D*e/(1-e*sin(A))
- Directrix at bottom side. Principal axis is y-axis. Focus F is origion.
- F is bottom side focus of x^2/a^2 + y^2/b^2 = 1.
- When A=270 we have D*e = (a-f)*(1+e)
R=D*e/(1+e*sin(A))
- Directrix at top of ellipse. Principal axis is y-axis. Focus G is origion.
- G is top side focus of x^2/a^2 + y^2/b^2 = 1.
- When A=90 we have D*e = (a-f)*(1+e)
Go to Begin
Q19. Quiz in ellipse
Quize and answer
Go to Begin
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