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janhardo

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11 years, 311 days
B. Ed math
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@vv ThanksOf course you must first ...

janhardo 910
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 910
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 910
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 910
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 910
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 910
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 910
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 910
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 910
July 10 2020
0 0

@Carl Love 

Thanks

The procedure conics must show this too the degenerate cases

@acer ThanksThis formula is suffici...

janhardo 910
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 910
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

@janhardo   .....

janhardo 910
July 09 2020
0 0

@janhardo 


 

> 

restart;
> 

B:=Array(1..5,1..2);

Matrix(5, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = 0, (3, 1) = 0, (3, 2) = 0, (4, 1) = 0, (4, 2) = 0, (5, 1) = 0, (5, 2) = 0})

 (1)
> 

T2:=Array([[0.5,-1],[1,2],[1.5,1],[1.75,2],[2,2.5]]);

Matrix(5, 2, {(1, 1) = .5, (1, 2) = -1, (2, 1) = 1, (2, 2) = 2, (3, 1) = 1.5, (3, 2) = 1, (4, 1) = 1.75, (4, 2) = 2, (5, 1) = 2, (5, 2) = 2.5})

 (2)
> 

op(eval(T2));

1 .. 5, 1 .. 2, {(1, 1) = .5, (1, 2) = -1, (2, 1) = 1, (2, 2) = 2, (3, 1) = 1.5, (3, 2) = 1, (4, 1) = 1.75, (4, 2) = 2, (5, 1) = 2, (5, 2) = 2.5}, datatype = anything, storage = rectangular, order = Fortran_order

 (3)
> 

op([2],T2); # go further with this

1 .. 5, 1 .. 2

 (4)

Can select all : 1..5 , 1 and 5  ;  1..2 , 1 , 2

> 

op([2,1],T2);

1 .. 5

 (5)
> 

op([2,1,1],T2);

1

 (6)
> 

op([2,1,2],T2);

5

 (7)
> 

op([2,2],T2);

1 .. 2

 (8)
> 

op([2,2,1],T2);

1

 (9)
> 

op([2,2,2],T2);

2

 (10)
> 

 

> 

#op([1],T2); nothing ,[2] is two-dimensional?

> 

op([2],T2);

1 .. 5, 1 .. 2

 (11)
> 

 

> 

 

> 

whattype(op([2],T2));

exprseq

 (12)
> 

upperbound(T2);

5, 2

 (13)
> 

lowerbound(T2);

1, 1

 (14)
> 

op([2,1,2], T2) ;

5

 (15)

 

> 

op(2, op(1, op(2, T2)));

Error, invalid input: op expects 1 or 2 arguments, but received 3

  

 

> 

 

> 

 


 

Download utzoeken_array_dimensie.mw

 

@Carl Love ThanksThe content of the...

janhardo 910
July 08 2020
0 0

@Carl Love 

Thanks

The content of the array and the array indexing  are related in this programming of "view" is that it? 
Its a way of modern programming then.

The wireframe ranges for domain must also be made general for any array procedure input

 

@acer ThanksCompared with the origi...

janhardo 910
July 08 2020
0 0

@acer 
Thanks

Looking to this explanation ...complicated

Compared with the original bookprogramming is this new programming more advanced , because the intervals for x, y and z  for the 3 axis by a given  input  array are also calculated  by the "view "code in the procedure.

Only how the view code this handles seems to be complicated. 

 

 

 

@acer  Thanks I agree with you, to...

janhardo 910
July 08 2020
0 0

@acer 

Thanks
I agree with you, to stay in a context for better understanding

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