Show that the conicoid S: x^2+2y^2+3z^2−2yz+4zx+6xy−2x−4y−6z+8=0 has a centre. Hence find the centre
If we have a conicoid, S’
ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0
then S' has a centre if the following set of equations has a solution
ax + hy + gz + u = 0
hx + by + fz + v = 0
gx + fy + cz + w = 0
(see: http://www.scribd.com/doc/25330499/Unit-1, section 1.4, Reduction to Standard Form)
In our conicoid S,
a = 1, b = 2, c = 3
f = -1, g = 2, h = 3
u = -1, v = -2, w = -3
giving our system of simultaneous equations as,
x + 3y + 2z - 1 = 0
3x + 2y – z – 2 = 0
2x – y + 3z – 3 = 0
The solution to this system of equations is
x = 19/21, y = -4/21, z = 1/3
Since we have found a valid solution, then there is a centre and that centre is at (19/21, -4/21, 1/3)
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