Mathematics Dictionary

Defintion
- 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.
The equation of the locus is (x-h)^2/a^2 + (y-k)^2/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.

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Q2. Equations of ellipse in rectangular form
A2. Answers

[Standard Form]

Equation : (x-h)^2/a^2 + (y-k)^2/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)^2/a^2 + (y-k)^2/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)^2/5^2 + (y-3)^2/3^2 = 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)^2/3^2 + (y-3)^2/5^2 = 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)^2/5^2 + (y+2)^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 Figure 2 Locus in polar coordinate

[Function 1]

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 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)^2/a^2 + y^2/b^2 = 1

R = D*e/(1-e*cos(A)) : Origin at F.
(x-f)^2/a^2 + y^2/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
A4. Answer

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)^2/a^2 + (y-k)^2/b^2 = cos(t)^2 + sin(t)^2.
- Since cos(t)^2 + sin(t)^2 = 1.
- Hence
- (x-h)^2/a^2 + (y-k)^2/b^2 = 1.

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Q5. Example : 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)^2/a^2 + (y-k)^2/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^2/5^2 + y^2/3^2 = 1.

[Method 2] By using distance formula

Since PF + PG = 10.
Sqr((x+4)^2 + y^2) + Sqr((x-4)^2+ y^2) = 10.
Square both sides we have
- (x+4)^2+y^2 + (x-4)^+y^2 + 2*Sqr((x+4)^2+y^2)*((x-4)^2+y^2)) = 100.
- 2*x^2+2*y^2+32 - 100 = 2*Sqr(((x+4)^2+y^2)*((x-4)^2+y^2)).
- x^2 + y^2 - 34 = Sqr(((x+4)^2+y^2)*((x-4)^2+y^2)).
Square both sides
- (x^2+y^2)^2 - 68*(x^2+y^2) + 34^2 = ((x+4)^2+y^2)*((x-4)^2+y^2).
- (x^2+y^2)^2 - 68*(x^2+y^2) + 34^2.
- = (x2-4^2)^2 + (y^2)*((x-4)^2 + (x^2+4)^2) + y^4.
Simplify and we get
- -68*x^2 - 68*y^2 +34^2 = - 32*x^2 + 32*y^2 + 16^2.
- -36*x^2 - 100*y^2 = 16^2 - 34^2.
- 36*x^2 + 100*y^2 = 900.
- x^2/5^2 + y^2/3^2 = 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)^2/3^2 + (y+3)^2/5^2 = 1, find coordinates of foci

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).

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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

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. Example : Convert R = 1.8/(1-0.8*cos(A) to x^2/a^2 + y^2/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 :
(x-1)^2/5^2 + (y-2)^2/3^2 = 1.
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 = negative.

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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)).
- R = (a-f) and R = D*e/(1+e) when A = 180 degrees and cos(A) = -1.
- 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)^2/a^2 + y^2/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
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Q15. Example : Draw tangent to ellipse using reflection
Figure 5 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.

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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).

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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

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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)

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Q19. Quiz in ellipse

1. Find the locus of the following equations :

a. (x-2)^2/5^2 + (y+3)^2/3^2 = 1.
b. 9*x^2 + 25*y^2 + 54*x + 100*y + 181 = 0
c. 9*x^2 + 25*y^2 + 54*x + 100*y - 44 = 0
d. 9*x^2 + 25*y^2 + 54*x + 100*y + 200 = 0
e. 9*x^2 + 20*x*y + 25*y^2 + 18*x + 100*y - 200 = 0

2. Compare (x-2)^2/5^2 + (y+3)^2/3^2 = 1 with (x-2)^2/3^2 + (y+3)^2/4^2 = 1.

a. Principal axis.
b. Focal length f.
c. Coordinate of Foci.
d. Vertice on principal axis
e. Polar form
f. Equation of directrix = 0

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Q20. Answer to Quiz in ellipse
1. Find the locus of the following equations :

a. (x-2)^2/5^2 + (y+3)^2/3^2 = 1.
- It is an ellipse.
- Principal axis y = -3.
b. 9*x^2 + 25*y^2 + 54*x + 100*y + 181 = 0
- 9*(x+3)^2 - 81 + 25*(y+4)^2 -100 + 181 = 0.
- 9*(x+3)^2 + 25*(y+2)^2 = 0.
- It is a point.
- Principal axis y = -3.
c. 9*x^2 + 25*y^2 + 54*x + 100*y - 116 = 0
- 9*(x+3)^2 - 81 + 25*(y+4)^2 - 100 - 44 = 0.
- 9*(x+3)^2 + 25*(y+2)^2 = 225.
- (x+3)^2/25 + (y+2)^2/9 = 1
- It is an ellipse.
- Principal axis y = -2.
d. 9*x^2 + 25*y^2 + 18*x + 100*y + 200 = 0
- 9*(x+3)^2 - 81 + 25*(y+4)^2 - 100 + 200 = 0.
- 9*(x+3)^2 + 25*(y+2)^2 = -19.
- No locus in real number system.
e. 9*x^2 + 20*x*y + 25*y^2 + 18*x + 100*y - 200 = 0
- Since B^2 - 4*A*C = 20^2 - 4*9*25 = -500
- Hence it is an ellipse.
- Or a point.
- Or no locus in real system.

2. Compare (x-2)^2/5^2 + (y+3)^2/3^2 = 1 with (x-2)^2/3^2 + (y+3)^2/4^2 = 1.

a. Principal axis.
b. Focal length f.
c. Coordinate of Foci.
d. Vertice on principal axis
e. Polar form
f. Equation of directrix = 0

For (x-2)^2/5^2 + (y+3)^2/3^2 = 1.

Principal axis is y = - 3.
Focal length f = Sqr(a^2-b^2) = 4.
Center is at (2,-3).
Coordinate of foci are (-2,-3) and (6,-3).
Coordinate of vertice are (-3,-3) and (7,-3).
Polar form : R = D*e/(1-e*cos(A)) or R = D*e/(1+e*cos(A)).
Equation of directrix
- Since e = f/a = 4/5 = 0.8.
- R = a - f = D*e/(1+e) when A = 180 degrees.
- Hence D*e = (a-f)*(1+e) = (5-4)*(1+0.8) = 1.8
- Hence D = 1.8/e = 2.25.
- Equation of directrix at x = -3 - 2.25 = - 5.25 for focus (-3,-3).
- Equation of directrix at x = 7 + 2.25 = 9.25 for focus (7,-3).

For (x-2)^2/3^2 + (y+3)^2/5^2 = 1.

Principal axis is x = 2.
Focal length f = Sqr(a^2-b^2) = 4.
Center is at (2,-3).
Coordinate of foci are (2,-7) and (2,1).
Coordinate of vertice are (2,-8) and (2,2).
Polar form : R = D*e/(1-e*sin(A)) or R = D*e/(1+e*sin(A)).
Equation of directrix
- Since e = f/a = 4/5 = 0.8.
- R = a - f = D*e/(1+e) when A = 270 degrees.
- Hence D*e = (a-f)*(1+e) = (5-4)*(1+0.8) = 1.8
- Hence D = 1.8/e = 2.25.
- Equation of directrix at y = -7 - 2.25 = - 9.25 for focus (2,-7).
- Equation of directrix at x = 1 + 2.25 = 3.25 for focus (2,1).

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