Show that the conicoid x^2..... (Analytical Geometry)

Question

Show that the conicoid x^2+2y^2+3z^2−2yz+4zx+6xy−2x−4y−6z+8=0 has a centre. Hence find the centre

Answer 1

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)

 

Karan, I don't know if you want/need to show the above link in your submission. I have included it for your own interest.

Comments

Thanks for the answer... I've attempted the question last night but got a different solution.... Maybe coz the values of f, g, h, u, v, w are totally different, and I disagree with your values.... For eg.:- the st. eq. says ax^2+.....+2Fyz+....+ d = 0.... Here coefficient of yz is 2F = (-2), so F = (-1).... And the same thing applies with the rest...

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Thanks for the answer... I've attempted the question last night but got a different solution.... Maybe coz the values of f, g, h, u, v, w are totally different, and I disagree with your values.... For eg.:- the st. eq. says ax^2+.....+2Fyz+....+ d = 0.... Here coefficient of yz is 2F = (-2), so F = (-1).... And the same thing applies with the rest... I you can understand what I mean here and so maybe our answers are different....

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Sorry, my bad. I doubled/misread the values for f,g,h,u,v,w

re-evaluated the system of equations.

Is that what you got?

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Well the centre I got was (1, (-2/7), 6)

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Your x- and y-coordinates are of the same order as mine.

I plotted this hyperboloid in Maple. I read from the 3d plot and got (very) rough approximations for the centre coordinates as (1, 0, 0.5)

Those values (roughly) confirm my own calculations.

Your z-coordinate looks a bit high ?

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Well then I must cross check