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janhardo

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@tomleslie  Thanks. I really thin...

janhardo 700
July 15 2020
0 0

@tomleslie 

Thanks.

I really think that there was such a toolbox focussed on programming in Maple back in 2000
Not focussed on Matlab.
Is there now a toolbox to get what supports programming?

On the other hand why it should not exists anymore, if there was such a toolbox in the past ?


 

@janhardo The domain roster is corr...

janhardo 700
July 14 2020
0 0

@janhardo

 The domain roster is correct, but not all planes can be used as input

 

 

@janhardo Got also matrix and deter...

janhardo 700
July 13 2020
0 0

@janhardo 
Also got education(not extensively)  in matrix and determinants theory,although a long time back , so i can study the subject,but it is a lot to study. 

https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections

The Ax^2 + By^2 + Cx + Dy + E = 0 general formula is programmed for conics by inspecting the formula and rearrange it and look at the conics definitions.   
Don't know if this is intendend to do it on this way by the authors of the book to improve your programming /analyzing skills at that time 2002 ( further in the book the authors using matrix and determinants fo rprogramming)

So i could be still possible to program conics2 by inspecting the general formula  Ax^2 + Cy^2 + Bxy + Dx + Ey + F = 0  and rearrange it .. ? ( see my post ) 

@vv Thanks!Looks interesting the pr...

janhardo 700
July 13 2020
0 0

@vv 

Thanks!

Looks interesting the programming and the result of course.
No background study material of this all, so there is nothing to learn for me from this as you mentioned earlier.

The original question was :

(2)  Write a conics2 procedure that will classify a quadratic equation of the form
             "Ax^2 + Cy^2 + Bxy + Dx + Ey + F = 0"


where there is a cross term. You might want to review the conic sections chapter in you calculus book.  This classification involves a rotation of axes and is considerably more complicated.

This means the programming of the book task: conics2 is not so advanced as your example

Maybe got the students for the programming course more background information than i had ?  
Its mission impossible for me

The nice thing on this procedure P is that it include also the conics from general formula
Ax^2 + By^2 + Cx + Dy + E = 0
It are parabolas, ellipses and hyperbolas in any orientation covered by  proc P and the degenerated cases

 I discovered in my programmed procedure conics : conics(1,2,3,4,5) gives "no graph"

that's wrong..no still correct!.
How could i ever doubt to this programming made by a professional    
P(1,0,2,3,4,5);
         2      2                      
        x  + 2 y  + 3 x + 4 y + 5 = 0, 

          Type = "This equation has no real solutions."  =. no graph


Is it that the rotation axes also must be plotted against the xoy system for the conics in the task ? 

. 
 

@vv ThanksOf course you must first ...

janhardo 700
July 13 2020
0 0

@vv 

Thanks

Yes, i agree with you
Of course you must first master the math if you want to optimal work/solve with it in Maple.
Doing it by hand first with exercises.
I did it with series : a topic not mastered by me and then in Maple makes it harder.

In this case of conics:=proc(A,B,C,D,E) it was only elementair formula manipulation for the general formule Ax^2 + By^2 + Cx + Dy + E = 0.

Note: 

example 

conics(1,2,3,4,5); ( all + ) 
                           "no graph" => wrong : something went wrong in the bookexample code ?.NO its correct

The same can be done for conics2 ,but is more complicated with all those cases to reckon with.

@vv ThanksI am aware of this , but ...

janhardo 700
July 13 2020
0 0

@vv 

Thanks

I am aware of this , but the exercise is about classifying conics it is not about solving for values
What i did for Ax^2 + By^2 + Cx + Dy + E = 0

You can try out ..

example 

conics(1,2,3,4,5); ( all + ) 
                           "no graph" => wrong : something wrong in the bookexample code 

conics(0,1,2,3,4); 
                       " laying parabola"

conics(0,0,2,3,4);
                             "line"

conics(0,0,0,3,4);
                             "line"

conics(0,0,0,0,4);
                           "no graph"

conics(0,0,0,0,0);
                       "the whole plane"

restart: with(student):with(plottools): with(plots):
conics:=proc(A,B,C,D,E)
  local N,M,O, ans;
    if A<>0 and B<>0 then #(1)
         M:=evalf((B*C^2+A*D^2-4*E*A*B)/(4*A*B)); 
           if M=0 then 
                 if A*B>0 then ans:= "point"
                 else ans:="two lines"
                 end if;
           elif M<>0 then
              if (M/A)>0 and (M/B)>0 then  ans:= "ellipse" 
             elif (M/A)<0 and (M/B)<0 then ans:= "no graph" 
             else ans:="hyperbola" 
             end if;
           end if;
     elif A=0 and B<>0 #(2)
           then N:=-E/B+D^2/(4*B^2);
            if C<>0 then ans:=" laying parabola"
            elif C=0 then 
              if N>0 then ans:="two lines" 
              elif N=0 then ans:="point" 
                else  ans:="no graph"
              end if;
            end if;
     elif A<>0 and B=0 #3 was excercise
           then O:=-E/A-C^2/(4*A^2);
            if D<>0 then ans:=" vertical parabola"
            elif D= 0 then 
              if O>0 then ans:="two lines" 
              elif O= 0 then ans:="point" 
                else  ans:="no graph"
              end if;
            end if;

     elif A=0 and B=0 then #(4)
           if C<>0 or D<>0 then ans:="line"
           elif E=0 then ans:="the whole plane"
           else ans:="no graph" 
           end if;
      end if;
      ans; 
     end proc:
------------------------------------------------------------

Ax^2 + Cy^2 + Bxy + Dx + Ey + F = 0 i try to classyfying conics for this 

It seems to be not yet correct this formulamanipulation above ?

Corrected    ...

janhardo 700
July 13 2020
0 0

Corrected 


 

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)
> 

 

> 

 

> 

 

> 

 

> 

 

> 

 

lijkt niet op eq1 !

(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

> 

 

> 

 

> 

restart;# via forum
> 

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

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

 (3)
> 

# complete squares in x and y, then separate the term with x or y

temp := selectremove(has,Student:-Precalculus:-CompleteSquare(f,[x,y]),{x,y});

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

 (4)
> 

# equate the first to the negation of the second

temp[1]=-temp[2];

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

 (5)
> 

# alternate way

((a,b)->a=-b)(temp);

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

 (6)
> 

# one way to treat the new rhs (there are others)

((a,b)->a=(numer/denom)(-b))(temp);

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

 (7)
> 

# with some forcing sorting of terms

((a,b)->sort(a,order=plex(A,B))=sort(numer(-b),order=plex(C,D))/denom(-b))(temp);

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

 (8)
> 

simplify( (8), 'symbolic' );

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

 (9)
> 

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

(1/16)*(4*A^2*x^2+((4*B*y+4*D)*x+4*C*y^2+4*E*y)*A+D^2)/(A^2*(D^2-4*F*A)) = 1

 (10)
> 

 

> 

 

Check-----------------------------------------------------------------------------------------------

> 

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

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

 (11)
> 

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

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

 (12)
> 

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

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

 (13)
> 

 

> 

 

> 

 

> 

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

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

 (14)
> 

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

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

 (15)
> 

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

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

 (16)
> 

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

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

 (17)
> 

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

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

 (18)
> 

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

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

 (19)
> 

((x + (B*y + D)/(2*A))^2)/C;

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

 (20)
> 

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

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

 (21)
> 

 

 


 

Download vraag1-herleidingconics2_1.mw

@janhardo The definitive procedure&...

janhardo 700
July 11 2020
0 0

@janhardo 

The definitive procedure


 

> 

restart; with(plots):
> 

conics:=proc(A,B,C,D,E)
> 

  local N,M,O, ans;
> 

    if A<>0 and B<>0 then #(1)
> 

         M:=evalf((B*C^2+A*D^2-4*E*A*B)/(4*A*B));

> 

           if M=0 then

> 

                 if A*B>0 then ans:= "point"
> 

                 else ans:="two lines"
> 

                 end if;
> 

           elif M<>0 then
> 

              if (M/A)>0 and (M/B)>0 then  ans:= "ellipse"

> 

             elif (M/A)<0 and (M/B)<0 then ans:= "no graph"

> 

             else ans:="hyperbola"

> 

             end if;
> 

           end if;
> 

     elif A=0 and B<>0 #(2)
> 

           then N:=-E/B+D^2/(4*B^2);
> 

            if C<>0 then ans:=" laying parabola"
> 

            elif C=0 then

> 

              if N>0 then ans:="two lines"

> 

              elif N=0 then ans:="point"

> 

                else  ans:="no graph"
> 

              end if;
> 

            end if;
> 

     elif A<>0 and B=0 #3 was excercise
           then O:=-E/A-C^2/(4*A^2);
> 

            if D<>0 then ans:=" vertical parabola"
> 

            elif D= 0 then

> 

              if O>0 then ans:="two lines"

> 

              elif O= 0 then ans:="point"

> 

                else  ans:="no graph"
> 

              end if;
> 

            end if;
> 

     elif A=0 and B=0 then #(4)
> 

           if C<>0 or D<>0 then ans:="line"
> 

           elif E=0 then ans:="the whole plane"
> 

           else ans:="no graph"

> 

           end if;
> 

      end if;
> 

      ans;

> 

     end proc:
> 

maplemint(conics);
> 

 

Input for conics : A,B,C,D,E to check : 0 and  +  
For 0 and 1  input gives 2^5 = 32 inputs for parabola, lines,points, hyperbola , whole plane, nothing.
Should be useful to check all input automated , but how?

Lets look for the

> 

 # ans:= conics(0, A, D, C, E).
> 

conics(2,0,5,1,-7)# y=-2x^2+5x-7 ;conics(0,0,1,2,3);#y= -0.5x-1.5

" vertical parabola"

  

"line"

 (1)
> 

display(implicitplot({2*x^2-5*x+y+7=0,x+2*y+3=0},x=-4..4,y=-8..4));

 

CDExercises

(1) Adjust the conics procedure to classify a circle as a circle instead of as an ellipse.

 

(2)  Write a conics2 procedure that will classify a quadratic equation of the form

Ax^2+Cy^2+Bxy+Dx+Ey+F = 0

 

where there is a cross term. You might want to review the conic sections chapter in you calculus book.  This classification involves a rotation of axes and is considerably more complicated.

Qestion: what is rotation of axes ? and what kind of graphs ?

 

-----------------------------------------------

Note : is conics() procedure above

the input to the procedure are coefficients of a quadratic equation of the form

"Ax^2+By^2+Cx+Dy+E=0 ---------------------"

 

 

> 

 


 

Download def_conics_procedure.mw

 

@janhardo  Made a analyse analog i...

janhardo 700
July 11 2020
0 0

@janhardo 

Made a analyse analog in the book for case 3

There was enough formula manipulation to do in Maple , but don't know exactly the handling of this all.
So i did it with drag and drop and as start with Student[Precalculus][CompleteSquare]( );

For case #3 (a is not 0 and B=0 ), the parabola is in normal position  


 

Classification of conic sections

mijn herleiding i als acer maar met drag and drop formule gedeelts

> 

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; # via contextmenu :

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

 (2)
> 

Student[Precalculus][CompleteSquare]( (2), [y] ); ## voor [x] en [y] tegelijk: zie commando van acer  # hier 1 voor  1 , eerst naar y

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

 (3)
> 

Student[Precalculus][CompleteSquare]( (2), [x] ); # via x  

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

 (4)
> 

 

> 

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

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

 (5)
> 

Student[Precalculus][CompleteSquare]( (5), [y] );

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

 (6)
> 

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

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

 (7)
> 

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

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

 (8)
> 

 

> 

 

=====================================================

case 2 A=0 en B niet 0

> 

 

> 

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

 

> 

 

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

 (9)
> 

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

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

 (10)
> 

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

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

 (11)
> 

 

> 

(-C*(x - E/C + D^2/(4*B*C)))/B;

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

 (12)
> 

simplify(%);

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

 (13)
> 

N=%; #

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

 (14)
> 

N= (-E/B)+(D^2/(4*B^2)); # for C = 0 in (14), N can be 0, + , or  -

N = -E/B+(1/4)*D^2/B^2

 (15)

================================================
case 3 A niet 0 en B=0

> 

 

> 

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

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

 (16)
> 

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

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

 (17)
> 

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

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

 (18)
> 

 

> 

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

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

 (19)
> 

-D*(y - E/D + C^2/(4*A*D))/A;

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

 (20)
> 

simplify(%);

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

 (21)
> 

O=%;

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

 (22)
> 

O=((-4*0*y + 4*E)*A - C^2)/(4*A^2);# for D=0 in( #22)

O = (1/4)*(4*A*E-C^2)/A^2

 (23)
> 

O= E/A-C^2/(4*A^2); # can be 0 , + , or -

O = E/A-(1/4)*C^2/A^2

 (24)
> 

 


 

Download vraag_herleiding_conic_sections_formule.mw

@janhardo Indeed a parabola in stan...

janhardo 700
July 10 2020
0 0

@janhardo 

Indeed a parabola in standard position
A error ..division by 0  


 

> 

restart; with(plots):
> 

conics:=proc(A,B,C,D,E)
> 

  local N,M,ans;
> 

    if A<>0 and B<>0 then #(1)
> 

         M:=evalf((B*C^2+A*D^2-4*E*A*B)/(4*A*B));

> 

           if M=0 then

> 

                 if A*B>0 then ans:= "point"
> 

                 else ans:="two lines"
> 

                 end if;
> 

           elif M<>0 then
> 

              if (M/A)>0 and (M/B)>0 then  ans:="ellipse"

> 

             elif (M/A)<0 and (M/B)<0 then ans:="no graph"

> 

             else ans:="hyperbola"

> 

             end if;
> 

           end if;
> 

     elif A=0 and B<>0 #(2)
> 

           then N:=-E/B+D^2/(4*B^2);
> 

            if C<>0 then ans:="parabola"
> 

            elif C=0 then

> 

              if N>0 then ans:="two lines"

> 

              elif N=0 then ans:="point"

> 

                else  ans:="no graph"
> 

              end if;
> 

            end if;
> 

     elif A<>0 and B=0 #3
           then N:=-E/B+D^2/(4*B^2);
> 

            if C<>0 then ans:="parabola"
> 

            elif C=0 then

> 

              if N>0 then ans:="two lines"

> 

              elif N=0 then ans:="point"

> 

                else  ans:="no graph"
> 

              end if;
> 

            end if;
> 

     elif A=0 and B=0 then #(4)
> 

           if C<>0 or D<>0 then ans:="line"
> 

           elif E=0 then ans:="the whole plane"
> 

           else ans:="no graph"

> 

           end if;
> 

      end if;
> 

      ans;

> 

     end proc:
> 

 conics(1,-1,-6,5,3);   

"hyperbola"

 (1)
> 

 

 

 
> 

conics(2,0,5,1,-7);

Error, (in conics) numeric exception: division by zero

  
> 

implicitplot(2*x^2-5*x+y+7=0,x=-4..4,y=-8..4);

 
> 

 


 

Download antw_symmetry_of_cases.mw

@Carl Love  Thanks That's a i...

janhardo 700
July 10 2020
0 0

@Carl Love 

Thanks

That's a ingenious way to look at this problem!
Then in case # 2 (A =0) and B not 0 ) the parabola is laying one
For case #3 (a is not 0 and B=0 ), the parabola is in normal position  

All other in # case 2 stays the same in #case 3
Try this out with the code 

@acer  Thanks   ...

janhardo 700
July 10 2020
0 0

@acer 

Thanks


 

> 

 

> 

restart;

The following procedure is for classifying conic sections and starts on page 84 of the text.   The input to the procedure are coefficients of a quadratic equation of the form

Ax^2+By^2+Cx+Dy+E = 0

and the ouput is the type of conic section the equation describes.

 

> 

conics:=proc(A,B,C,D,E)
> 

  local N,M,ans;
> 

    if A<>0 and B<>0 then #(1)
> 

         M:=evalf((B*C^2+A*D^2-4*E*A*B)/(4*A*B));

> 

           if M=0 then

> 

                 if A*B>0 then ans:= "point"
> 

                 else ans:="two lines"
> 

                 end if;
> 

           elif M<>0 then
> 

              if (M/A)>0 and (M/B)>0 then  ans:="ellipse"

> 

             elif (M/A)<0 and (M/B)<0 then ans:="no graph"

> 

             else ans:="hyperbola"

> 

             end if;
> 

           end if;
> 

     elif A=0 and B<>0 #(2)
> 

           then N:=-E/B+D^2/(4*B^2);
> 

            if C<>0 then ans:="parabola"
> 

            elif C=0 then

> 

              if N>0 then ans:="two lines"

> 

              elif N=0 then ans:="point"

> 

                else  ans:="no graph"
> 

              end if;
> 

            end if;
> 

      elif A<>0 and B=0  then ans:="Exercise" #(3)EXERCISE
> 

      elif A=0 and B=0 then #(4)
> 

           if C<>0 or D<>0 then ans:="line"
> 

           elif E=0 then ans:="the whole plane"
> 

           else ans:="no graph"

> 

           end if;
> 

      end if;
> 

      ans;

> 

     end proc:
> 

 

> 

 

Example 1

Let's test the procedure on a few equations.  We will start with ones we can classify on sight.  For instance the circle  x^2+y^2 = 4 .

 

> 

conics(1,1,0,0,-4);

"ellipse"

 (1)

 

If we look back at the code we see that this is the correct output.  The procedure classifies a circle as an ellipse whose foci coincide.  Let's try another example.

Example 2

> 

conics(2,3,1,5,-7);

"ellipse"

 (2)

This corresponds to the equation

2*x^2+3*y^2+x+5*y-7 = 0

We can visually check this by looking at the graph.

 

> 

with(plots):
> 

implicitplot(2*x^2+3*y^2+x+5*y-7=0,x=-4..4,y=-5..5,grid=[75,75]):

 

In the procedure conics as extra to add a small plot as output next to the classification name .  
For case # 3 ( A ≠ 0 o B= 0 )  EXERCISE we need a derived formula based on (1.6) -section start formula

``

> 

 

start formula

   

For case # 3 ( A ≠ 0 o B= 0 )

> 

 

> 

((a,b)->sort(a,order=plex(A,B))=sort(numer(-b),order=plex(C,D))/denom(-b))(temp);

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

 (3)

from this formula (3) to this formula below  for further analyse
How to this with Maple?

> 

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

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

 


 

Download vraag_2_herleiding_conic_sections_formule.mw

@Carl Love  Thanks The procedure ...

janhardo 700
July 10 2020
0 0

@Carl Love 

Thanks

The procedure conics must show this too the degenerate cases

@acer ThanksThis formula is suffici...

janhardo 700
July 09 2020
0 0

@acer 

Thanks

This formula is good enough.

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

> 

 

The whole analyse starts with

A*x^2+B*y^2+C*x+D*y+E= 0 
 It defines , conics , lines, point, plane , nothing ,  ?

It are preparations for a new procedure for classifying conic sections 
At page 83 of the enclosed pdf  at number 1 : it starts with a characterization

 

@acer ThanksLooks the correct formu...

janhardo 700
July 09 2020
0 0

@acer 

Thanks

Looks the correct formula, yes and the command used to produce this is not simple.

With order of addens you mean : leftside of equation : starts with A ?
That doesn't matter here, although the book starts with A( )^2 +.B( )^2 

M is now the new name of rhs of the equation

M = (BC^2+AD^2-4EAB)/4AB shows the book

Maple can do it all, but it is specialized use

I must investigate your derived formula for different conditions for M(=0) and A and B for sign combinations.
It seems to be quicker done by hand, then using Maple for this ?   
blz83.pdf

blz84.pdf

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