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SIMPLIFICATION OF THE GENERAL EQUATION OF THE SECOND DEGREE BY TRANSLATION

Let us explore the question of what translation of the coordinate system might reduce the general equation of the second degree

1) f(x, y, z) = ax^{2} + by^{2} + cz^{2} + 2fyz + 2gxz + 2hxy + 2px + 2qy + 2rz + d = 0

to simpler form. More specifically, let us ask what translation might cause the first degree terms to disappear. Let us observe the result of substituting

x = x' + x_{0}

y = y' + y_{0}

z = z' + z_{0}

into equation 1) above.

The equation becomes

2) f(x', y', z') = f(x' + x_{0}, y' + y_{0}, z' + z_{0}) = a(x' + x_{0})^{2} + b(y' + y_{0})^{2} + c(z' + c_{0})^{2} + f(y' + y_{0})(z'
+ z_{0}) + 2g (x' + x_{0})(z' + z_{0}) + 2h(x' + x_{0})(y' + y_{0}) + 2 p(x' + x_{0}) + 2q (y' + y_{0}) + 2r(z' + z_{0}) + d

= ax'^{ 2} + by'^{ 2} + cz'^{ 2} + 2fy'z' + 2gx'z' + 2hx'y' + 2(ax_{0} + hy_{0} + gz_{0} + p)x' + 2(hx_{0} + by_{0} + fz_{0}
+ q)y' + 2(gx_{0} + fy_{0} + cz_{0} + r)z' + ax_{0}^{2} + by_{0}^{2} + cz_{0}^{2} + 2f y_{0}z_{0} + 2gx_{0}z_{0} + 2hx_{0}y_{0} + 2px_{0} + 2qy_{0} +
2rz_{0} + d

and, finally

3) f(x', y', z') = ax'^{ 2} + by'^{ 2} + cz'^{ 2} + 2fy'z' + 2gx'z' + 2hx'y' + 2(ax_{0} + hy_{0} + gz_{0} + p)x' +
2(hx_{0} + by_{0} + fz_{0} + q)y' + 2(gx_{0} + fy_{0} + cz_{0} + r)z + f(x_{0}, y_{0}, z_{0}) = 0

We see from 3) above that all the coefficients a, b, c, f, g, h of the second degree terms remain
unchanged under the translation. For the first degree terms to disappear we can see that the
coefficients 2(ax_{0} + hy_{0} + gz_{0} + p), 2(hx_{0} + by_{0} + fz_{0} + q), and 2(gx_{0} + fy_{0} + cz_{0} + r) of x', y' and
z' must be zero. Thus the first degree terms should be eliminated by a translation to that (x_{0}, y_{0},
z_{0}) satisfied by the following set of equations:

ax_{0} + hy_{0} + gz_{0} + p = 0

4) hx_{0} + by_{0} + fz_{0} + q = 0

gx_{0} + fy_{0} + cz_{0} + r = 0

Let us note that this set is precisely the set of equations defining a center of a quadric surface.
Thus we have shown that the quadric surface centers do represent points at which the first degree
terms disappear. So if we do a translation to a center the second degree terms remain the same
and the first degree terms disappear. Now let us ask what the constant term d becomes on a
translation to a center. Let d' denote the constant term at a center. From 3) above we see that d'
= f(x_{0}, y_{0}, z_{0}). What is the value of f(x_{0}, y_{0}, z_{0}) at a center? The function

f(x_{0}, y_{0}, z_{0}) = ax_{0}^{2} + by_{0}^{2} + cz_{0}^{2} + 2fy_{0}z_{0} + 2gx_{0}z_{0} + 2hx_{0}y_{0} + 2px_{0} + 2qy_{0} + 2rz_{0} + d

can be written as

f(x_{0}, y_{0}, z_{0}) = (ax_{0} + hy_{0} + gz_{0} + p)x_{0} + (hx_{0} + by_{0} + fz_{0} + q)y_{0} + (gx_{0} + fy_{0} + cz_{0} + r)z_{0} + (px_{0} + qy_{0}
+ rz_{0} + d)

and we can see from inspection of this equation and 4) above that the first three terms vanish at a
center (x_{0}, y_{0}, z_{0}). Thus the new constant term has the value

d' = f(x_{0}, y_{0}, z_{0}) = px_{0} + qy_{0} + rz_{0} + d

Theorem. If a translation is made to a center (x_{0}, y_{0}, z_{0}) of the quadric surface

f(x, y, z) = ax^{2} + by^{2} + cz^{2} + 2fyz + 2gxz + 2hxy + 2px + 2qy + 2rz + d = 0

the transformed equation becomes

f(x', y', z') = ax'^{ 2} + by'^{ 2} + cz'^{ 2} + 2fy'z' + 2gx'z' + 2hx'y' + px_{0} + qy_{0} + rz_{0} + d = 0

Theorem. The equation obtained from that of a given quadric surface by a translation to a center of the surface is independent of the center chosen.

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