Mathematics Dictionary

From oxy to uov by rotating angle A
- | +cos(A) +sin(A) |
- | -sin(A) +cos(A) |
Express u and v in terms of x and y
- u = +x*cos(A) + y*sin(A)
- v = -x*sin(A) + y*cos(A)

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AN 14 02. Convert x*y = 1 to standard hypebola form

Method 1 : Rotating 45 degrees

u = +x*cos(45) + y*sin(45)
v = -x*sin(45) + y*cos(45)
Since cos(45) = sin(45) = Sqr(2)/2
Hence we have
- u =(Sqr(2)/2)*(+x + y)
- v =(Sqr(2)/2)*(-x + y)
Solve for x and y
- x = (u - v)/(Sqr(2))
- y = (u + v)/(Sqr(2))
Substitue x and y into x*y = 1, we have (u^2 - v^2) = -2
The hyperbola is (u/Sqr(2))^2 - (v/Sqr(2))^2 = -1
- The principal axis is y = x
- The semi-axis : a = Sqr(2) and b = Sqr(2)
- The center : (0, 0)
- The vertex : (-1, -1) and (1, 1) in oxy system
- The vertex : (-Sqr(2), 0) and (+Sqr(2), 0) in ouv system
- The focal length is f = Sqr(a^2 + b^2) = 2

Method 2 : By diagram

We know that it is hyperbola because B^2 - 4*A*C = (1) - 4*0*0 = 1
From diagram we know the principal axis is y = x
From diagram we know the center to vertex (-1, -1) or (1, 1) is Sqr(2)
Hence the semi-axis : a = Sqr(2) and b = Sqr(2)
Let y = x be the ou of the ouv system
Then the equation is according to rotating angle
- (u/Sqr(2))^2 - v/Sqr(2))^2 = +1
- (u/Sqr(2))^2 - v/Sqr(2))^2 = -1

Questions : From AN 08

Find the coordinate of foci
Find the equation of directrix

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AN 14 03. Convert (x + 1)*(y - 2) = 1 to standard hyperbola form
Method

Use translation to change above equation to x*y = 1
Use AN 14 02, to get the standard form (u/a)^2 - (v/b)^2 = 1
Change it as ((u - h)/a)^2 - ((v - k)/b)^2 = 1 by translation
Where h = -1 and k = +2

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AN 14 04. Eliminate x*y in conic section

F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

Question
- Eliminate x*y term by rotating an angle
- The angle is tan(Ang) = (A - C)/B
After rotation : F(,u,v) = P*u^2 + Q*u*v + R*v^2 +S*u + T*v + W = 0
- Rotation : u = x*cos(Ang) + y*sin(Ang) and v = -x*sin(Ang) + y*cos(Ang)
- Substitute rotation formula into F(x,y) = 0 and simplify we have
  - P = A * Cos(ANG) ^ 2 + C * Sin(ANG) ^ 2 + B * Cos(ANG) * Sin(ANG)
  - Q = 2*(C - A)*cos(Ang)*sin(Ang) + B*(cos(Ang)^2 - sin(ANg)^2)
  - R = A * Sin(ANG) ^ 2 + C * Cos(ANG) ^ 2 - B * Cos(ANG) * Sin(ANG)
  - S = D * Cos(ANG) + E * Sin(ANG)
  - T = -D * Sin(ANG) + E * Cos(ANG)
  - W = F
To elliminate x*y term means Q = 0 in F(u, v)
- Hence 2*(A - C)*cos(Ang)*sin(Ang) = B*(cos(Ang)^2 - sin(Ang)^2)
- Hence (A - C)*sin(2*Ang) = B*cos(2*Ang)
- Hence tan(2*Ang) = (B)/(A - C)
or Ang = 0.5*arctan(B/(A - C))

Example : Rotate 45 degrees of x*y = 1

A = 0, B = 1, C = 0, D = 0, E = 0, F = -1
Hence tan(2*Ang) = (B)/(A - C) = infinite and Ang = pi/2
Hence Ang = 45 degrees
P = +B*cos(ang)*sin(Ang) = 1/2
R = -B*cos(ANg)*sin(Ang) = -1/2
S = 0 and T = 0
W = -1
After rotating 45 degree, the new equation (u^2)/2 - (v^2)/2 - 1 = 0
Standard hyperbola form is (u/Sqr(2))^2 - (v/Sqr(2))^2 = 1

Computer method

Numerical calculator Program NC 04 01 : elliminate x*y
Data : 0,1,0,0,0,-1

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AN 14 05. Rotate 45 degrees of x^2 - y^2 = 1

u = Sqr(2)*(x + y)/2
v = Sqr(2)*(-x + y)/2
Hence x = Sqr(2)*(u - v)/2
Hence y = Sqr(2)*(u + v)/2
Substitute x and y into x^2 - y^2 = 1
((u - v)^2)/2 - ((u + v)^2)/2 = 1
Simplify we have u*v = -1/2

Computer method

Numerical calculator Program NC 04 03 : Rotate 45 degrees
Data : 1,1,0,0,45

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AN 14 06. Sketch F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

Method

1. Eliminate x*y term by rotating angle ang = 0.5*arctan(B/(A-C))
2. Use new coefficients and completing square to get standard form
3. Use standard form to plot in oxy system
3. Rotate the plot by angle ang = 0.5*arctan(B/(A-C))

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AN 14 07. Rotate the graphs of functions of conic sections

Computer method

Numerical calculator AN 21 14
Example : Rotate 45 degrees of x*y = 1

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AN 14 08. Draw circum-circle of triangle ABC

Case 1

If we know the radius and center is related to one side
This side is x-axis, we can sketch the circle

Case 2

If no side is along x-axis, the center and radius are related with one side
To plot the circle, we have to use rotating techniques
Example : Circom-center and radius are related with angle A of triangle
- Radius : R = a/(2*sin(A))
- Center I to side AB is IF = Sqr(R^2 - (AB/2)^2)
- Hence center related to AB is (AB/2, IF)
- If AB is x-axis, we can use coodinate of center and radius to draw circle
- If AB makes angle Z with x-axis, how to draw on computer ?
- We use rotation technique
Rotation angle U
- Let cneter be (h,k) in oxy system
- Let center be (u,v) in ouv system after rotating U
  - u = +h*cos(U) + k*sin(U)
  - v = -h*sin(U) + k*cos(U)
- Now we can use center (u,v) and radius R to draw the circle
- The circle will pass the three vertices of triangle ABC.

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AN 14 09. Draw in-circle of triangle ABC

Center and radius related angle A and side AB

Tanget from A to in-circle is (s - a) where s = (a+b+c)/2
Radius : r = (s-a)*tan(A/2)
Hence center is h = (s-a) and k = r
If AB makes angle U with x-axis,
The center should be changed as
- u = +h*cos(U) + k*sin(U)
- v = -h*sin(U) + k*cos(U)
Now we can use (u,v) as center to draw the circle
The three sides of triangle ABC will tangent the in-circle

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AN 14 10. Draw ex-circle of triangle ABC

Center and radius related angle A and side AB

Tanget from A to in-circle is s where s = (a+b+c)/2
Radius : r = s*tan(A/2)
Hence center is h = s and k = r
If AB makes angle U with x-axis,
The center should be changed as
- u = +h*cos(U) + k*sin(U)
- v = -h*sin(U) + k*cos(U)
Now we can use (u,v) as center to draw the circle
The three sides of triangle ABC will tangent the ex-circle

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AN 14 00. Outlines

Rotation matrix

From oxy to uov by rotating angle
- | +cos(A) +sin(A) |
- | -sin(A) +cos(A) |
Express u and v in terms of x and y
- u = +x*cos(A) + y*sin(A)
- v = -x*sin(A) + y*cos(A)

Elliminate x*y in F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

Change F(x,y) to F(u,v) by totating
After rotation : F(,u,v) = P*u^2 + Q*u*v + R*v^2 +S*u + T*v + W = 0
- P = A * Cos(ANG) ^ 2 + C * Sin(ANG) ^ 2 + B * Cos(ANG) * Sin(ANG)
- Q = 2*(C - A)*cos(Ang)*sin(Ang) + B*(cos(Ang)^2 - sin(ANg)^2)
- R = A * Sin(ANG) ^ 2 + C * Cos(ANG) ^ 2 - B * Cos(ANG) * Sin(ANG)
- S = D * Cos(ANG) + E * Sin(ANG)
- T = -D * Sin(ANG) + E * Cos(ANG)
- W = F
To elliminate x*y term means Q = 0 in F(u, v)
- Hence 2*(A - C)*cos(Ang)*sin(Ang) = B*(cos(Ang)^2 - sin(Ang)^2)
- Hence (A - C)*sin(2*Ang) = B*cos(2*Ang)
- Hence tan(2*Ang) = (B)/(A - C) or Ang = 0.5*arctan(B/(A - C))
The principal axis is Ou or y = s*x where s = tan(Ang)

Function y = 1/x is hyperbola

Since x*y = 1 is hyperbola
Rotate 45 degrees of x*y = 1, it becomes (x/a)^2 - (y/b)^2 = 1

Graphs of x*y in F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

If A = C and B = 0, it is a circle
If B^2 - 4*A*C = 0, it is a parabola
If B^2 - 4*A*C LT 0, it is an ellipse
If B^2 - 4*A*C GT 0, it is a hyperbola

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