##Sage, 247 bytes
Sage, 247 bytes
def f(p):
for i in[1,2,4]:
z=[i&1,i&2,i&4,0,0,0]
M=matrix([z]+[[x*x,x*y,y*y,x,y,1]for x,y in p])
try:A,B,C=(M\vector(z))[:3]
except:continue
d=B*B-4*A*C
return['parabola','hyperbola','circle','ellipse'][[d==0,d>0,d<0and B==0and A==C,d<0].index(1)]
This function takes an iterable of (x,y) pairs as input, tries computing the discriminant of each of the 3 possible linear systems (A=1, B=1, and C=1), and outputs the type of conic section based on the values of the discriminant, A, B, and C.
There's probably some more golfing to be done, but I'm rusty with Sage and sleepy right now, so I'll work on it more in the morning.
##Sage, 247 bytes
def f(p):
for i in[1,2,4]:
z=[i&1,i&2,i&4,0,0,0]
M=matrix([z]+[[x*x,x*y,y*y,x,y,1]for x,y in p])
try:A,B,C=(M\vector(z))[:3]
except:continue
d=B*B-4*A*C
return['parabola','hyperbola','circle','ellipse'][[d==0,d>0,d<0and B==0and A==C,d<0].index(1)]
This function takes an iterable of (x,y) pairs as input, tries computing the discriminant of each of the 3 possible linear systems (A=1, B=1, and C=1), and outputs the type of conic section based on the values of the discriminant, A, B, and C.
There's probably some more golfing to be done, but I'm rusty with Sage and sleepy right now, so I'll work on it more in the morning.
Sage, 247 bytes
def f(p):
for i in[1,2,4]:
z=[i&1,i&2,i&4,0,0,0]
M=matrix([z]+[[x*x,x*y,y*y,x,y,1]for x,y in p])
try:A,B,C=(M\vector(z))[:3]
except:continue
d=B*B-4*A*C
return['parabola','hyperbola','circle','ellipse'][[d==0,d>0,d<0and B==0and A==C,d<0].index(1)]
This function takes an iterable of (x,y) pairs as input, tries computing the discriminant of each of the 3 possible linear systems (A=1, B=1, and C=1), and outputs the type of conic section based on the values of the discriminant, A, B, and C.
There's probably some more golfing to be done, but I'm rusty with Sage and sleepy right now, so I'll work on it more in the morning.
##Sage, 247 bytes
def f(p):
for i in[1,2,4]:
z=[i&1,i&2,i&4,0,0,0]
M=matrix([z]+[[x*x,x*y,y*y,x,y,1]for x,y in p])
try:A,B,C=(M\vector(z))[:3]
except:continue
d=B*B-4*A*C
return['parabola','hyperbola','circle','ellipse'][[d==0,d>0,d<0and B==0and A==C,d<0].index(1)]
This function takes an iterable of (x,y) pairs as input, tries computing the discriminant of each of the 3 possible linear systems (A=1, B=1, and C=1), and outputs the type of conic section based on the values of the discriminant, A, B, and C.
There's probably some more golfing to be done, but I'm rusty with Sage and sleepy right now, so I'll work on it more in the morning.