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

```