## How to build a hyperbola...

I know how to contruct an isoscele ABCD trapeze knowing the 2L length of AB and BC=CD=CD=a. But I don't know answer to that question : L being fixed together with points A and B , show that the place of C when a varies is a branch of hyberbola. Here is my code.

restart; unprotect(O, D);
with(plots):
Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end:
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end:
_EnvHorizontalName := x: _EnvVerticalName := y:
O:=[0,0]:A:=[-L,0]:alpha:=arccos((2*L-a)^2/(2*a*(2*L-a))):h:=tan(alpha)*(2*L-a)/2:
B:=[L,0]:C:=[L-(2*L-a)/2,h]:D:=[-L+(2*L-a)/2,h]:
L := 6; a := 7;
poly := [A, B, C, D, A];tp := textplot([[A[], "A"], [B[], "B"], [C[], "C"], [D[], "D"]], color = black, 'align' = {'above', 'right'});
trapeze := polygonplot(poly, axes = normal, color = "DarkGreen", transparency = .9);
display([tp, trapeze], scaling = constrained); Thank you foryour help.

## calcula tion of an angle...

We consider an ellipse  E defined by the equation x²/a²+y²/b²=1. Either F1 and F2 its foci.
We consider a point M of E  such the angle F1,M,F2 is equal to a given value phi.
Find F1M et F2M in function of a, b and phi . Then find the coordiinates x0 and y0 of M in function of a, b, and phi. Here is the begining of my code;

restart;
_EnvHorizontalName := x: _EnvVerticalName := y:
with(plots):
line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:
Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end proc:
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end proc:

ell := x^2/a^2+y^2/b^2 = 1:
c := sqrt(a^2-b^2):
F1 := [c, 0]: F2 := [-c, 0]:
M := [a*cos(t), b*sin(t)]:
MF2 := simplify(expand(sqrt((c+a*cos(t))^2+(b*sin(t))^2)));
MF1 := simplify(expand(sqrt((c-a*cos(t))^2+(b*sin(t))^2)));
varphi := arccos((MF1^2+MF2^2-4*c^2)/(2*MF1*MF2));
f := unapply(varphi, t);
`max&varphi;` := f((1/2)*Pi);
diff(f(t), t);

a := 7: b := 5: t := (1/3)*Pi:
evalf(varphi);
evalf(MF1); evalf(dist(M, F1));
evalf(MF2); evalf(dist(M, F2));
MF1+MF2; evalf(%);
`max&varphi;`; evalf(%);
ELL := implicitplot(ell, x = -a-2 .. a+2, y = -2-b .. b+2, color = blue):
tp := textplot([[F1[], "F1"],[F2[],"F2"],[M[],"M"]], 'align' = {'above', 'left'}):
po := plot([F1,F2, M], style = point, symbolsize = 15, symbol = solidcircle, color = red):

display([ELL, tp, po], scaling = constrained): Would  you help me to answer the questions ? Thank you.

## solve nonlinear integral equation...

Hi. i have a nonlinear integral equation. any idea for solving it? tnx in advance

integral-equation.mw

## How to simplify the trigonometric expression...

How to simplify this trigonometric expression with Maple? I only know the way with  identify command, which is difficult to call mathematically correct:

```restart;
Expr:=arctan((1-tan(20*Pi/180))/(1-tan(25*Pi/180)));
evalf[15](Expr);
identify(%);```

## How to write on a drawing P1, P1 P3......

How to write names on an drawing P1, P2, P3, Pn... I try
textplot({seq([(coordinates(point(P[i]))), cat("P", i)], i = 0 .. 5)}, 'align' = {'above', 'left'}):;
i got the message : wrong number of arguments that i don't how correct. Thank you.

## hexagon inscribed and exinscrit in a circle...

restart; with(plots):unprotect(gamma):
_EnvHorizontalName := 'x';_EnvVerticalName := 'y';
line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:
R := 3:
ang := [0, (1/3)*Pi, 3*Pi*(1/4)+.2, 7*Pi*(1/6)+.4, 8*Pi*(1/5), 13*Pi*(1/7)]:
for i to 6 do P || i := [R*cos(ang[i]), R*sin(ang[i])] end do:
pts := [seq(P || i, i = 1 .. 6)]:
for i to 6 do tang || i := x*P || i[1]+y*P || i[2] = R^2 end do:
sol:=solve({tang1,tang3},{x,y}):Q13:=[subs(sol,x),subs(sol,y)]:
sol := solve({tang1, tang2}, {x, y}): Q1 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang2, tang3}, {x, y}): Q2 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang3, tang4}, {x, y}): Q3 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang4, tang5}, {x, y}): Q4 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang5, tang6}, {x, y}): Q5 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang1, tang6}, {x, y}): Q6 := [subs(sol, x), subs(sol, y)]:
ptQ := [seq(Q || i, i = 1 .. 6)]:
line14 := line(Q1[1], Q1[2], Q4[1], Q4[2]): L14 := implicitplot(line14, x = -10 .. 10, y = -10 .. 10, color = red):
line25 := line(Q2[1], Q2[2], Q5[1], Q5[2]): L25 := implicitplot(line25, x = -10 .. 10, y = -10 .. 10, color = red):
line36 := line(Q3[1], Q3[2], Q6[1], Q6[2]): L36 := implicitplot(line36, x = -10 .. 10, y = -10 .. 10, color = red):
sol := solve({line14, line25}, {x, y}): I1 := [subs(sol, x), subs(sol, y)]:

lineP23 := line(P2[1], P2[2], P3[1], P3[2]): lineP56 := line(P5[1], P5[2], P6[1], P6[2]):
sol := solve({lineP23, lineP56}, {x, y}): gamma := [subs(sol, x), subs(sol, y)]:
lineP12 := line(P1[1], P1[2], P2[1], P2[2]): lineP45 := line(P4[1], P4[2], P5[1], P5[2]):
sol := solve({lineP12, lineP45}, {x, y}): beta := [subs(sol, x), subs(sol, y)]:
lineP34 := line(P3[1], P3[2], P4[1], P4[2]): lineP16 := line(P1[1], P1[2], P6[1], P6[2]):
sol := solve({lineP16, lineP34}, {x, y}): alpha := [subs(sol, x), subs(sol, y)]:
pl:= line(alpha[1], alpha[2], gamma[1], gamma[2]):
hexa := seq(implicitplot(tang||i, x = -20 .. 20, y = -20 .. 20, linestyle=3,color = blue),i=1..6):
#hexa:=plot([seq([P||i,P||(i mod 6)+1],i=1..6),color=green):
#hexa:=plot([seq([P||i,P||(i+1)],i=1..6)],thickness=4,color=green):
hex:=plot([[Q1,Q2],[Q2,Q3],[Q3,Q4],[Q4,Q5],[Q5,Q6],[Q6,Q1]],thickness=4,color=green):
tp := textplot({seq([op(pts[i]), cat("P", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
tpq := textplot({seq([op(ptQ[i]), cat("Q", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
TP:=textplot([[I1[],"I"],[alpha[],"alpha"],[beta[],"beta"],[gamma[],"gamma"]],'align' = {'above', 'left'}):
slopes:=[seq(((dx,dy)->dy/dx)((pts[i]-pts[(i mod 6)+1])[]),i=1..6)]:
lines:=zip((pt,slope)->y=slope*(x-pt[1])+pt[2],pts,slopes):
plotpts:=plot(pts,style=point,colour=red,symbol=solidcircle,symbolsize=5):
plotptQ:=plot(ptQ,style=point,colour=black,symbol=solidcircle,symbolsize=5):
plotlines:=plot(rhs~(lines),style=line,linestyle=3,colour=magenta):
cir:=implicitplot(x^2+y^2=R^2,x=-R..R,y=-R..R,color=black):
PL:=implicitplot(pl,x=-1..15,y=-15..6,color=blue,thickness=3):
display(plotpts,plotptQ,plotlines,hex,cir,L14,L25,L36,PL,tp,tpq,TP,axis = [gridlines = [4, color = blue]],
view=[-6..10,-15..6],scaling=constrained,axes=none,size=[800,800]):
How to simplify this program ? Thank you.

## How to make a drawing according to special rules i...

Hello. I have certain requirements for drawings. Please tell me, is it possible to implement them when plotting in Maple? Thanks.
Drawing width 15 cm
Be sure to frame and mesh. The thickness of the frame and serifs on the axes is 0.5 pt. Mesh thickness 0.25 pt. The length of the serifs is 1.2 mm.
The thickness of the graph itself is 1 pt
Axis names 10 pt

## Solution-Of-Second-Order-Differential-Equation...

How to show that the function z0(t)=t²+exp(-t)  is solution of z"(t)+2z'(t)+z(t)=2exp(-t); Thank you.

## Labels on the axes...

Hello. Tell me, please, is it possible to somehow remove the labels on the axes in Maple when plotting graphs? I mean labels 1, 2, 3 and so on. Thank you for your help.

## How to manage homogenous coordinates...

If the equation with homogeous coordinates in the base (A,B,C) is pYZ+qZX+rXY=0, an affine  equation in the base (A,AB,AC) is
pxy+qy(1-x-y)+r(1-x-y)=0, it is also written rx²+(q+r-p)xy+qy²=0. The discrim of  trinom is rx²+(q+r-p)xy+qy² is of the sign of
delta=(q+r-p)²-4qr; We get a hyperbole, a parabole, an ellipse according to delta is >0, =0 or <0 respectvely. How implement
this property. Thank you very much.

## Simplifying expressions with sqrt...

The problem: to simplify the expression

for any negative  x  and  y .

Below we see that Maple copes with the task brilliantly (example 1). For example, it presents  sqrt(x*y)  as  sqrt(-x)*sqrt(-y)  and so on. But the same technique, applied only to the numerator of this expression does not give the desired presentation in the form of a square (example 2 and example 3).

```restart;
# Example 1
A:=(x+y-2*sqrt(x*y))/(sqrt(-x)+sqrt(-y));
simplify(A) assuming negative;
factor(%,{sqrt(-x),sqrt(-y)});
```

```restart;
# Example 2
B:=x+y-2*sqrt(x*y);
simplify(B) assuming negative;
factor(%,{sqrt(-x),sqrt(-y)});
```

```restart;
# Example 3
B:=x+y-2*sqrt(x*y);
R:=simplify(B) assuming positive;
combine(R) assuming positive;
factor(R,{sqrt(x),sqrt(y)});
```

Two questions:

1. Does anyone know the reasons for this behavior.

2. Does anyone know an easy way to simplify in examples 2 and 3 (without  substitutions  like  x=+-u^2  and  y=+-v^2 and so on,  of course)

## Conics through 3 points with homogeous coordinates...

An exercise of 1948 commits me to form the equation of conics passing by 3 points. Let P=0, Q=0, R=0 be the equations of the sides of the triangle ABC; if we associate these sides 2 to 2 we obtain 3 conics passing through points A, B, C having for equations QR=0, RP=0, PQ=0. As a result, the general equation of conics around the triangle ABC is: aQR+bRP+cPQ=0. P, Q, R being equations of the form mx+ny+pz=0 (so-called homogeneous coordinates). Then change to refined coordinates with x+y+z=1 (formula found on the internet and surely misinterpreted). Is it necessary to change of base ? Thank you for your help.

## Serious bugs in solve command

by: Maple 2018

In the two examples below (in the second example, the range for the roots is simply expanded), we see bugs in both examples (Maple 2018.2). I wonder if these errors are fixed in Maple 2020?

 > restart;
 > solve({log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2,x>=-7*Pi/2,x<=-2*Pi}, explicit, allsolutions); # Example 1 - strange error message solve({log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2,x>=-4*Pi,x<=-2*Pi}, explicit, allsolutions);  # Example 2 - two roots missing
 (1)
 > plot(log[1/3](2*sin(x)^2-3*cos(2*x)+6)+2, x=-7*Pi/2..-2*Pi); plot(log[1/3](2*sin(x)^2-3*cos(2*x)+6)+2, x=-4*Pi..-2*Pi);
 > Student:-Calculus1:-Roots(log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2, x=-7*Pi/2..-2*Pi);  # OK Student:-Calculus1:-Roots(log[1/3](2*sin(x)^2-3*cos(2*x)+6)=-2, x=-4*Pi..-2*Pi);  # OK
 (2)
 >

I am glad that  Student:-Calculus1:-Roots  command successfully handles both examples.