Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Conic Section : Hyperbola

Questions

Symbol Definition

Example : x^2 = square of x

Q01 | - Locus of a hyperbola in rectangular coordinates

Q02 | - Equations of hyperbola in rectangular coordinates

Q03 | - Equations or hyperbola in polar form

Q04 | - Equations of hyperbola in parametric form

Q05 | - Find hyperbola if focus F(2,0), vertex U(0,0) and P(2,6) are given

Q06 | - Hyperbola : (x-2)^2/3^2 - (y+3)^2/4^2 = 1, find coordinates of foci

Q07 | - Hyperbola : Focus F(0,0), directrix x=-D. Find equation if e=2

Q08 | - Convert x^2/4^2 - y^2/3^2 = 1 to polar form

Q09 | - Convert R = 6/(1-2*cos(A)) to (x-h)^2/a^2 - y^2/b^2 = 1

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

Q11 | - Formula

Q12 | - Reference

Q13 | - Prove the locus of hyperbola is x^2/a^2 - y^2/b^2 = 1

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

Q15 | - Draw tangent to hyperbola by law of reflextion

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

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

Q18 | - Compare polar form with standard rectangular form

Q19 | - Compare x^2/a^2 - y^2/b^2 = 1 with x^2/a2 - y^2/b^2 = -1

Answers

Q00. Graphs of hyperbola

Diagrams about hyperbola

Graphs in Analytic geometry Section 8 in anlytic geometry
Figures
- 08 01 ((x-h)/a)^2 - ((y-k)^2) = 1
- 08 02 ((x-h)/a)^2 - ((y-k)^2) = -1
- 08 03 R = D*e/(1 - e*sin(A))
- 08 04 R = D*e/(1 + e*sin(A))
- 08 05 R = D*e/(1 - e*cos(A))
- 08 06 R = D*e/(1 + e*cos(A))
- 08 13 ((x-h)/4)^2 - ((y-k)/3)^2 = 1
- 08 14 ((x-h)/4)^2 - ((y-k)/3)^2 = -1
- 08 15 R = D*e/(1 - e*cos(A))
- 08 16 R = D*e/(1 + e*cos(A))
Note : e GT 1

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Q01. Locus of a hyperbola in rectangular coordinates
Defintion

Figure 1
Two fixed points F and G. A moving point P(x,y).
When P moves so that |PF| - |PG| = constant = 2*a, what is the locus ?
Answer : The locus of point P is a hyperbola.
Defintion
- Where F and G are the foci. The value a is the major semi-axis.
- The foci F and G are on the principal axis.
- The equation is
  - The center of the hyperbola is C(h,k).
  - (x-h)^2/a^2 - (y-k)^2/b^2 = +1 if the princial axis in x-axis direction.
  - (x-h)^2/a^2 - (y-k)^2/b^2 = -1 if the princial axis in y-axis direction.
- Focal length : f = CF = CG = focal length = Sqr(a^2+b^2).
- Vertices on principal axis : CU = a and CV = a.
- Distance between focus and vertex is FU = f - a.

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Q02. Equations of hyperbola

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.
- The principal axis is along x-axis
- The principal axis is y = k.
- The foci F and G are on principal axis.
- The vertice U and V are on pricipal axis.
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.
- The principal axis is along y-axis
- The principal axis is x = h.
- The foci F and G are on principal axis.
- The vertice U and V are on pricipal axis.
The focal length is CF or CG which is Sqr(a^2+b^2).
The vertice
- The end points alnog principal axis is U and V. The a = CU = CV.
- FU = CF - CU = f - a.

Implicit Form without x*y

Equation : F(x,y) = A*x^2 + C*y^2 + D*x + E*y + F = 0.
Hyperbola : A and C must have different signs.
Princial axis is parallel to x-axis or y-axis.
Foci are on princial axis.
Find foci, a, b and f of hyperbola.
- Change to standard form by using completing the square.
- The standard form is (x-h)^2/a^2 - (y-k)^2/b^2 = 1.
- Hence we can find the center (h,k) and semi-axis a and b.
- Hence we can find f using f = Sqr(a^2+b^2).

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 a hyperbola.
The pricipal axis is y = m*x + n and 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 by using 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)^2/4^2 - (y-3)^2/3^2 = 1

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

[Example] Study (x+2)^2/3^2 - (y-3)^2/4^2 = -1

Pricipal axis is x = -2.
Center at (-2,3), a=4 and b=3.
Focal length f = Sqr(a^2+b^2) = Sqr(4^2+3^2) = 5.
Foci 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 a hyperbola.
- 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) = Sqr(34).
- Foci at x = h - f, y = k and x = h + f, y = k.

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Q03. Equations in Polar Form

Figure 2 Locus in polar coordinate

Function 1

Function : R = D*e/(1-e*sin(A)).
Origin is at F and directrix is y = -D.
D is focus to directrix line and e is eccentricy greater than 1.
A is angle making with the x-axis.
Foci are on x-axis.

Function 2

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

Function 3

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

Function 4

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

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

R = D*e/(1-e*cos(A)) : Origin at F and e greater than 1.
(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 = f-a when A = 180 degrees.
Hence D*e = (f-a)*(1+e).

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

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.
Locus of P is hyperbola if PF/PQ = e and greater than 1.
Where PF = R and PQ = D+x.
D = distance from F to directrix.
Prove that R = D*e/(1-e*cos(A)).
- R/PQ = e.
- 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 in parametric form
Equation 1

x = h + a*sec(t).
y = k + b*tan(t).
(h,k) are center. The values of a and b are semi-axis.
It is hyperbola since (x-h)^2/a^2 - (y-k)^2/b^2 = 1.

Equation 2

x = h + a*tan(t)
y = k + b*sec(t)

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Q5. Example : Find equation hyperbola if 3 points are given

Give focus F(2,0) and vertix U(0,0).
A point is P(2,6) on the hyperbola.
Find equation of the hyperbola.

Method 1

Construction
- Draw a horizon line as principal axis (x-axis).
- Draw U(0,0) as origin and vertex of the hyperbola.
- Draw focus F(2,0) on x axis.
- Draw point P(2,6)
Find the equation of the parabola.
- Let the center of parabola be C(h,0).
- Then the equation is (x-h)^2/a^2 - y^2/b^2 = 1.
- Find h, f, a and b
  - CU = a, CF = f and h = -a.
  - Then UF = CF - CU = f - a = 2 or f = 2 + a.
  - It is known that b^2 = f^2 - a^2 = (2+a)^2 - a^2.
  - Or b^2 = 4*(1+a).
- Substitue h, b^2 into above we have (x+a)^2/a^2 - y^2/(4+4*a) = 1.
- Substitue x=2 and y=6 into equation we have (2+a)/a^2 - 36/(4+4*a) = 1.
- Solve for a and we get a = 2.
- Hence center is at x = -2. Then f = 4.
- And b^2 = f^2 - a^2 = 4^2 - 2^2 = 12
- Hence equation is (x+2)^2/4 - y^2/12 = 1.

Method 2 : Polar equation

Draw rectangular coordinates system.
Translate above points to left by 2 units.
- That is F(0,0), U(-2,0) and P(0,6).
- Let Q be point on diretrix and PQ be perpendicular to directrix.
- Then PF = 6 and UF = 2.
Then polar equation is R = D*e/(1-e*cos(A)).
Find D*e.
- If A = 90 degrees, cos(A) = 0. R = PF = 6.
- Then 6 = D*e/(1+0) or D*e = 6.
Find e
- If A = 180 degrees, cos(A) = -1. R = UF = 2.
- Then 2 = D*e/(1+e) or 2*(1+e) = D*e = 6.
- Hence e = 2.
Hence R = 6/(1-2*cos(A))

Method 3 : Conver polar equation to (x-h)^2/a^2 - (y-k)^2/b^2 = 1

Find h, a, k, b.
- From method 2 we have e = 2 and D = 3.
- Also center C(h,k) where h = -a and k = 0.
- CF = f and CU = a.
Find a and f
- Since e = f/a = 2 or f = 2*a.
- When A = 180 degrees, R = UF = 2 = CF - CU = f - a.
- Hence f = 2 + a or 2*a = 2 + a.
- Hence a = 2 and f = 4.
Find b^2
- Since b^2 = f^2 - a^2 = 4^2 - 2^2 = 16 - 4 = 12.
Substitute a, b, h and k into equation.
Hence equation is (x+2)^2/4 - y^2/12 = 1.

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Q6. Example : If (x-2)^2/3^2 - (y+3)^2/4^2 = 1, find coordinates of foci

Principal axis is y = -3.
The foci are on principal axis.
The center is at C(2,-3).
Focal length f = Sqr(3^2 + 4^2) = 5.
Coordinate of focus is at F(2,-8).
Coordinate of other focus is at G(2,2).

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Q07. Hyperbola : Focus F(0,0), directrix x = -3. Find equation if e = 2

[Method 1] Polat form is R = D*e/(1-e*cos(A)).

D = distance of F to directrix = 3.
The eccentricity is e = 2.
Hence the polar equation is R = 6/(1-2*cos(A)).

[Method 2] The rectangular form is (x-h)^2/a^2 - y^2/b^2 = 1

Since F is the origin, and the principal axis is y = 0.
Hence the center is C(h,0).
Let U be the vertex and then CU = a and CF = f.
Also h = -f.
Find relation of f and a.
- UF = f - a.
- When A = 180, R = UF = D*e/(1+e).
- Hence D*e = (1+e)*(f-a).
- Hence 6 = (1+2)*(f-a) or f-a = 2.
- e = f/a and f = e*a = 2*a.
- Hence a = 2 and f = 4
Find b
- Since f = Sqr(a^2+b^2) or b^2 = f^2 - a^2.
- b^2 = 4^2 - 2^2 = 12.
The equation is (x+4)^2/2^2 - y^2/12 = 1.

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Q8. Example : Convert x^2/4^2 - y^2/3^2 = 1 to polar form

The polar form is R = D*e/(1-e*cos(A))
- F is origin
- Directrix is x = -D.
- When A=180 degrees, R = f-a.
- Hence f - a = D*e/(1+e) or D*e = (f-a)*(1+e).
x^2/4^2 - y^2/3^2 = 1
- Left focus is at (-5,0).
- Focal length f = Sqr(4^2+3^2) = 5.
- The eccentricity e = f/a = 5/4 = 1.25.
- D*e = (a-f)*(1+e) = (5-4)*(1+1.25) = 2.25.
Hence required polar function is R = 2.25/(1-1.25*cos(A))

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Q9. Example : Convert R = 2.25/(1-1.25*cos(A) to (x-h)^2/a^2 - y^2/b^2 = 1

R = D*e/(1-e*cos(A))
- D*e = 2.25 and e = 1.25.
- When A = 180, R = f-a = D*e/(1+e)
- Hence D*e = (f-a)*(1+e) = 2.25
- e = f/a = 1.25.
Find a
- Substitute f = 1.25*a into D*e.
- Hence (1.25*a-a)*(1+1.25) = 2.25.
- Hence 0.25*a = 1 and hence a = 4.
Find b
- f = 5 and a = 4.
- Since f = Sqr(a^2 + b^2) = 5 and hence b = 3.
The required equation is x^2/4^2 - y^2/3^2 = 1.

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Q10. Example : 9*x^2 - 16*y^2 - 18*x - 64*y - 71 = 0, find a, b ,f

Change it to standard form by completing the square.
- 9*(x^2 - 2x + 1 - 1) - 16*(y^2 + 4*y + 4 - 4) - 71 = 0.
- 9*(x-1)^2 - 9 + 25*(y+2)^2 - 64 - 71 = 0.
- 9*(x-1)^2 + 16*(y+2)^2 = 144.
Divide both sides by 144 and we have :
(x-1)^2/4^2 - (y+2)^2/3^2 = 1.
- Hence a = 4 and b = 3.
- Then f = Sqr(a^2+b^2) = 5.
- Principal axis is y = -2.
This is a hyperbola because (B^2 - 4*A*C) = 0 + 4*9*25 = positive.

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Q11. Formula :
(x-h)^2/a^2 - (y-k)^2/b^2 = 1

Definiton from construction
- Principal axis is y = k.
- Vertice on principal axis are at U and V.
- Foci on principal axis are at F and G.
- Center is at (h,k).
- CU = CV = a = Semi-axis on principal axis.
Focal length
- f = Sqr(a^2 + b^2).
- e = f/a.
- f = CF = CG
Efficiency e = f/a.
Locus in rectangular system |PF| - |PG| = 2*a.
- F and G are foci.
- P is moving point.
- Equation is (x-h)^2/a^2 - (y-k)^2/b^2 = 1
Locus in polar system R/PQ = e.
- P is moving point.
- PQ is distance from P to Q on directrix.
- F is origin and PF = R.
- Equation is R = D*e/(1-e*cos(A)) and e greater than 1.
Relation between polar and rectangular :
- R = D*e/(1-e*cos(A)).
- R = UF = (f-a) when A = 180 degrees.
- Use polar formula we have D*e = (f-a)*(1+e).
- D is the distance from focus F to directrix.
Slope of point on hyperbola dy/dx = (x*b^2)/(y*a^2).
Asymptotes
- y - k = +b*(x-h)/b
- y - k = -b*(x-h)/b

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Q12. References :

Sketch (x-h)^2/a^2 - (y-h)^2/b^2 = 1 ..... MD2002 ZM40 04

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 hyperbola in Index File
See contents of ZH.txt in Contents by chapres

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Q13. Example : Prove that locus of hyperbola 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 = f^2 - a^2.
- Hence b^2*x^2 - a^2*y^2 + 2*b^2*x*f = a^2*b^2 - b^2*f^2.
- (f^2-a^2)*x^2 - a^2*y*2 + 2*b^2*x*f = b^2*(f^2-a^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.
- R = b^2/(a - f*cos(A))
After elliminting f by f=e/a we have : R = b^2/(a - e*cos(A)/a).
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 = (f^2 - a^2)/a = (f-a)*(a+f)/a = (f-a)*(1+e) = D*e.
Hence R = D*e/(1-e*cos(A)).