Is it  possible to draw an ellipse inscribled in a given triangle and we know its orthoptist circle ? Thank you very much.

Etude d'un cas particulier a := 5: b := 7: k := 9: A := [a, 0]: B := [0, b]: #A et B fixes P := [t, 0]: Q := [0, k/t]:#P et Q 2 points mobiles cir := -a*x-b*y+x^2+y^2 = 0: sol := solve(subs(y = 5, cir), x): cen := [solve(diff(cir, x)), solve(diff(cir, y))]: x0 := sol[1]: y0 := 5: M := [x0, y0]: R := sqrt(cen[1]^2+cen[2]^2): beta := arctan(diff(solve(EQ(M, cen), y), x)): Recherche des valeurs de t pour que les 2 droites soient perpendiculaires eq := t^2*(y0-b)+t*(a*b-a*y0+b*x0-k)-x0*(a*b-k) = 0; sol := solve(eq, t); t := sol[1]; tp := sol[2]; P1 := [t, 0]; Q1 := [0, k/t]; PQ1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0:#1ere tangente PQ2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0:#2ième tangente P2 := [tp, 0]; Q2 := [0, k/tp]; CIR := implicitplot(cir, x = -4 .. 8, y = -4 .. 12, color = red); Fig := proc (alpha) local Dr1, DR1, Dr2, DR2, N, u0, v0, Po, t, tp, sol; global a, b, k, cen, R; u0 := cen[1]+R*cos(alpha); v0 := cen[2]+R*sin(alpha); N := [u0, v0]; sol := solve(t^2*(v0-b)+t*(b*u0-a*v0+a*b-k)-u0*(a*b-k) = 0, t); t := sol[1]; tp := sol[2]; Dr1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0; DR1 := implicitplot(Dr1, x = -4 .. 8, y = -4 .. 12, color = brown); Dr2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0; DR2 := implicitplot(Dr2, x = -4 .. 8, y = -4 .. 12, color = pink); Po := pointplot([N[]], symbol = solidcircle, color = [black], symbolsize = 8); display([Po, DR1, DR2]) end proc; DrPQ1 := implicitplot(PQ1, x = -4 .. 22, y = -4 .. 12, color = blue); DrPQ2 := implicitplot(PQ2, x = -4 .. 22, y = -4 .. 12, color = blue); Points := pointplot([A[], B[], M[], P1[], P2[], Q1[], Q2[], cen[]], symbol = solidcircle, color = [green], symbolsize = 10); T := plots:-textplot([[A[], "A"], [B[], "B"], [M[], "M"], [P1[], "P1"], [P2[], "P2"], [Q1[], "Q1"], [Q2[], "Q2"], [cen[], "cen"]], font = [times, 10], align = {below, left}); n := 19; display([seq(Fig(2*i*Pi/n), i = 0 .. n), Fig(beta), CIR, DrPQ1, DrPQ2, Points, T], scaling = constrained, size = [500, 500]); I would find out the focus of the ellipse. Thank you.

restart; with(plots):with(LinearAlgebra):unprotect(O); alias(conj = conjugate); conj z = lambda*v+a; La droite D est représentée par son équation complexe Appelons H l'affixe h le pied de la perpendicukaire abaissée de O sur (D) Les vecteurs OH et V sont orthogonaux donc z = lambda v + a h*conj(v)+conj(h)*v = 0; Le point H appartient à la droite (D) donc : h = lambda*v+a; conj(h) = conj(a)+lambda*conj(v); (lambda*v+a)*conj(v)+(conj(a)+lambda*conj(v))*v = 0; solve(%, lambda); h := simplify(subs(lambda = %, lambda*v+a)); a := 3-I*4; v := -2/3+4*I; evalc(h); H := [Re(h), Im(h)]; Représentation graphique d'un cas particulier : f := proc (x) options operator, arrow; -3*x+5 end proc: a := 3: A := [a, f(a)]:O:=[0,0]: zo := [8/3+I*f(8/3)]; ze := [2+I^(eval(diff(f(x), x), x = 2))]; Zo := [8/3, f(8/3)]; Ze := [2, f(2)]; ex := -3*x+5; V := `

conj := conjugate; d := a*x+b*y-c = 0; z := x+I*y; evalc(z+conj(z)); evalc(z-conj(z)); d := expand((1/2)*a*(z+conj(z))+b*(z-conj(z))/(2*I))-c; is(d = z(a-I*b)+conj(z)*(a+I*b)-2*c); varpi = a+I*b; is(d = z*conj(varpi)+conj(z)*varpi-2*c); How to perform calculations correctly ? Thank you.

z1 := a1+I*b1; z2 := a2+I*b2; abs(z1) = 1; abs(z2) = 1; argument(z1) = alpha; argument(z2) = beta; On considère dans ℂ les complexes z1 et z2 de module 1 et d'argument α et β Show that (z1+z2)^2/(z1+z2) est un réel positf ou nul. Dans quel cas est-il nul ? is((z1^2+2*z1*z2+z2^2)/(z1+z2) = z1/z2+z2/z1+2);#wrong answer z1/z2 = exp(I*(alpha-beta)); z2/z1 = exp(I*(beta-alpha)); is(z1/z2+z2/z1+2 = 2*(1+cos(alpha-beta)));#wong answer Miscalculations. Thank you for your help.