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Characterize the ellipsoid described by ...?
5 x^2 + 6 y^2 + 7 z^2 + 4 xy + 4 yz - 40 x - 80 y - 62 z + 37 = 0
Find it's center, and the length and orientation of its axes.
1 Answer
- RaffaeleLv 77 years agoFavorite Answer
center is (2,5,3)
semi-axes are 3, √6, √3
and axes have directions, respectively
(1,2,2),(-2,-1,2),(2,-2,1)
I procedeed in this way
first I found the center, via translation
x = X + n; y = Y + n; z = Z + p
5(X + m)^2 + 6(Y + n)^2 + 7(Z + p)^2 + 4(X + m)(Y + n) + 4(Y + n)(Z + p)
- 40(X + m) - 80(Y + n) - 62(Z + p) + 37 = 0
5X^2 + 4XY + 2X(5m + 2n - 20) + 6Y^2 + 4YZ + 4Y(m + 3n + p - 20) + 7Z^2
+ Z(4n + 14p - 62) + 5m^2 + m(4n - 40) + 6n^2 + n(4p - 80) + 7p^2 - 62p + 37 = 0
setting = 0 all the coefficient of first degree terms I got the system
5m + 2n - 20 = 0
m + 3n + p - 20 = 0
4n + 14p - 62 = 0
m = 2, n = 5, p = 3
so
C(2, 5, 3) is the center
then I looked for the 3x3 matrix A such that
([X;Y;Z] - [2;5;3])' A ([X;Y;Z] - [2;5;3]) = 1
see
http://en.wikipedia.org/wiki/Ellipsoid#Generalised...
and I found
A =
5 2 0
2 6 2
0 2 7
eigenvalues are 1/a^2, 1/b^2, 1/c^2
eigenvectors are the directions of the axis of the ellipsoid
therefore, in a suitable coordinate system (ξ, υ, ζ)
the canonical equation of the ellipsoid is
ξ²/9 + υ²/6 + ζ²/3 = 1