Ahhh...I haven't gotten used to noticing the **foo** being interpreted as bold
by my OutlookExpress. I see that the y*2 and e*2 should actually be y**2 and
e**2.
The comments I made still apply and the only difference is that the two values
of y should be substituted into eq3. Doing so yields (quickly) the following 4
solutions:
0.24915 0.294148
4.4 8.76118
0.24915 0.294148
4.4 8.76118
I see that there are duplicates so perhaps you can take either value of y to
generate the solution. If so, then what I said/showed before will still work:
just take y[0] and continue on with the solution process.
Also, I would recommend that you pass the parameters to hyperplot. I put
everything together below. Notice that your a2 are the numerical values, but
symbolically they are handled as na2, etc... in the general-solving part of the
routine.
def hyperplot(a2, b2, c2, d2, e2):
"""e2 is the distance between foci and a2,b2,c2,d2 are the times
of arrival the foci, c2 is at the center"""
a, b, c, e, na2, nb2, nc2, nd2, ne2 = symbols('a b c e na2 nb2 nc2 nd2 ne2')
r1=dict(
a = ((na2 - nc2)/2)**2,
b = ((nb2 - nc2)/2)**2,
c = ((nd2 - nc2)/2)**2,
e = ne2/2 )
r2=dict(
na2 = a2,
nb2 = b2,
nc2 = c2,
nd2 = d2,
ne2 = e2)
eq1 = y**2/(e**2 - a) - (x + e)**2/a + 1
eq3 = y**2/(e**2 - c) - (x - e)**2/c + 1
#fails
#return solve([eq1, eq3], [x, y])
ans = []
ys = solve(eq1, y)[0] # the two solutions yield duplicate so take first
xs = solve(eq3.subs(y, yi).as_numer_denom()[0], x)
# now make subs x->x2; x2->numerical x2
for xi in xs:
xi = xi.subs(r1).subs(r2).n()
yi = ys.subs(r1).subs(r2).subs(x, xi).n()
ans.append((xi, yi))
return ans
sol = hyperplot(
a2 = 5.10125066033543,
b2 = 1.844025414703168,
c2 = 2.303750131640994,
d2 = 0,
e2 = 3)
for s in sol:
print '%g %g' % s
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