Cerca di stabilire (senza computer) quali tipi di curve hanno le seguenti equazioni. Verifica la risposta
tracciandone il grafico: 8*(x-1)^2+y^2-6*y*(x-1)=0, (x-2)*y-1=0, x-y^2-2*y=0, x^2+x*y+y^2-1=0. |
source("http://macosa.dima.unige.it/r.R") PIANO(-4,6, -4,6) G = function(x,y) 8*(x-1)^2+y^2-6*y*(x-1); curva(G,"red") H = function(x,y) (x-2)*y-1; curva(H,"green3") K = function(x,y) x-y^2-2*y; curva(K,"blue") F = function(x,y) x^2+x*y+y^2-1; curva(F,"black") # # # # Risposta dettagliata usando il comando Conic (vedi) # Riscrivo i polinomi (qui uso WolframAlpha per far prima) # simplify 8*(x-1)^2 + y^2 - 6*y*(x-1) -> 8 x^2 - 6 x y - 16 x + y^2 + 6 y + 8 # BF=3; HF=3; PLANE(-4,6, -5,5) G = c(8,-6,1,-16,6,8); Conic(G); CURV(C1,"red") # C1=function(x,y) 8 * x^2 + -6 * x*y + 1 * y^2 + -16 * x + 6 * y + 8 # the conic is 2 lines # H = c(0,1,0,0,-2,-1); Conic(H); CURV(C1,"seagreen") # C1=function(x,y) 0 * x^2 + 1 * x*y + 0 * y^2 + 0 * x + -2 * y + -1 # equilateral hyperbola xC = 2 yC = 0 # dir.trasv.ax 0.5 0.5 dir.coniug.ax -0.5 0.5 # V1 3 1 V2 1 -1 # F1 3.414214 1.414214 F2 0.5857864 -1.414214 eccentr 1.414214 # In C2 there is the curve formed by the 2 asymptotes # K = c(0,0,-1,1,-2,0); Conic(K); CURV(C1,"blue") # C1=function(x,y) 0 * x^2 + 0 * x*y + -1 * y^2 + 1 * x + -2 * y + 0 # parabola xV = -1 yV = -1 dir.ax.symm = 1 0 # F -0.75 -1 In C2(x,y) there is the directrix # L = c(1,1,1,0,0,-1); Conic(L); CURV(C1,"black") # C1=function(x,y) 1 * x^2 + 1 * x*y + 1 * y^2 + 0 * x + 0 * y + -1 # ellipse xC = 0 yC = 0 # dir.major ax: 315 ° dir.minor ax: 45 ° # semi-major axis 1.414214 semi-minor axis 0.8164966 # F1 0.8164966 -0.8164966 F2 -0.8164966 0.8164966 eccentr 0.8164966 # Then you have the length of the string that draws the Ellipse with # StringE(xF1,yF1, xF2,yF2) [which is = major axis = distance V1-V2] # and the couple of hor/vert dist. from F1 (F2) to V1 (V2) with FV # # Le coniche con rappresentati alcuni degli elementi generati dal programma