janhardo

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11 years, 153 days

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These are questions asked by janhardo

How to install this old package for Maple 6 , should be trivial but turns te be out a puzzle with no result so far.

### This is the Maple initialization file created for use with the 
### LAMP modules. For Maple 6 in Windows 95, 98, 2000, or NT, this file 
### should be installed in the folder 
###
###        C:\Program Files\Maple 6\BIN.WNT
###
### The last three letters, WNT, may vary with the version of Windows.
###
### The following lines set up libname, so Maple knows to look in the
### proper folder to find the Lamp library. If you did not put the 
### Lamplib folder in C:\lamp6\lamplib, you must edit the first line.

lamppath := "C:\\lamp6\\lamplib\\":
libname := `lamppath`, libname:

### The following line enables one to use the command read(lampstart) 
### to establish the Lamp environment. It is optional.

lampstart := cat(lamppath,"lamp.txt"):

I am going now into 3D with the conics 
Ax^2+BY^2 +Cz^2 +Dxy'..........................................(general formula quadric surfaces)

Is there a existing procedure in Maple for this. 
Perhaps also with different coordinate systems ?
Probably is this too much asked.

I followed the post about two intersected cilinders.

 

  

Back in 2000 there was a Maple programming toolbox to buy 
Can't find it ?

Some rewriting for a general equation , how to do ?


 

Transformation of eq 1 in wanted eq 2 form

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

After complete square for x and y and some rearrangment (general form 2) ,  i got this equation eq1  

 

 

(A*(x + (B*y + D)/(2*A))^2 + C*(y + (B*x + E)/(2*C))^2)/(-C*y^2 - E*y - 2*F + (B*y + D)^2/(4*A) - A*x^2 - D*x + (B*x + E)^2/(4*C))=1;

(A*(x+(1/2)*(B*y+D)/A)^2+C*(y+(1/2)*(B*x+E)/C)^2)/(-C*y^2-E*y-2*F+(1/4)*(B*y+D)^2/A-A*x^2-D*x+(1/4)*(B*x+E)^2/C) = 1

(1)

eg1:= (-4*A^2*C*x^2 + (-4*y^2*C^2 + (-8*B*x*y - 4*D*x - 4*E*y)*C - (B*x + E)^2)*A - (B*y + D)^2*C)/(4*A^2*C*x^2 + (4*y^2*C^2 + (4*D*x + 4*E*y + 8*F)*C - (B*x + E)^2)*A - (B*y + D)^2*C) = 1;

(-4*A^2*C*x^2+(-4*y^2*C^2+(-8*B*x*y-4*D*x-4*E*y)*C-(B*x+E)^2)*A-(B*y+D)^2*C)/(4*A^2*C*x^2+(4*y^2*C^2+(4*D*x+4*E*y+8*F)*C-(B*x+E)^2)*A-(B*y+D)^2*C) = 1

(2)

 

(1) wanted form of eg1
"A x^2+B x y+C y^2+D x+E y+F=0*("general form 2))

for M ≠ 0

 

(x+(B*y+D)/(2*A))^2/X+  (y+(B*x+E)/(2*C))^2/Y = 1

 

Note : what X and Y could be ? :   M/A and M/B  ..

So it must be possible to transform eq1  in this form  above ?

 

Some background info

If B= 0 in (general form 2) you get (general form 1)  

 

(2) In book for M ≠ 0 as example we get this form
 Ax^2+Cy^2+Dx+Ey+F = 0(general form 1 )

(x+C/(2*A))^2*A/M+  (y+D/(2*B))^2*B/M = 1

 


 

Download vraag1-herleidingconics2.mw

 

 

I must do some formula manipulation
 

Classification of conic sections

restart; with(student):

A*x^2+B*y^2+C*x+D*y+E=0;

A*x^2+B*y^2+C*x+D*y+E = 0

(1)

f:= A*x^2+B*y^2+C*x+D*y+E;

A*x^2+B*y^2+C*x+D*y+E

(2)

completesquare(f, x);

A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y+E

(3)

f:= A*x^2+B*y^2+C*x+D*y+E;

A*x^2+B*y^2+C*x+D*y+E

(4)

ans:=Student[Precalculus][CompleteSquare]( (4), [y] );

B*(y+(1/2)*D/B)^2+A*x^2+C*x+E-(1/4)*D^2/B

(5)

ans1:=Student[Precalculus][CompleteSquare]( (4), [x] );

A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y+E

(6)

ans+ans1;

B*(y+(1/2)*D/B)^2+A*x^2+C*x+2*E-(1/4)*D^2/B+A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y

(7)

B*(y + D/(2*B))^2 + A*x^2 + C*x + 2*E - D^2/(4*B) + A*(x + C/(2*A))^2 - C^2/(4*A) + B*y^2 + D*y = 0;

B*(y+(1/2)*D/B)^2+A*x^2+C*x+2*E-(1/4)*D^2/B+A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y = 0

(8)

Now i must  get this form A( )^2 +B( )^2 = M  

 


 

Download vraag_herleiding_conic_sections_formule.mw

 

 

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