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SIMPLIFICATION OF THE GENERAL EQUATION OF THE SECOND DEGREE BY ROTATION AND TRANSLATION OF THE COORDINATE SYSTEM
Simplification by rotation of coordinate system. With a suitable rotation of the coordinate system the term in xy can be eliminated from any second degree equation. Under a rotation of the coordinate system about its origin by an angle of θ degrees the general equation of the second degree
f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
becomes
a'x' 2 + 2h'x'y' + b'y'2 + 2g'x' + 2f'y' + c' = 0
where
a' = a cos2 θ + 2 h sin θ cos θ + b sin2 θ
h' = (b - a) sin θ cos θ + h (cos2 θ - sin2 θ)
=
(b - a) sin 2θ + h cos 2θ
b' = a sin2 θ - 2h sin θ cos θ + b cos2 θ
g' = g cos θ + f sin θ
f' = f cos θ - g sin θ
c' = c
The condition that the xy term be eliminated by the rotation is that h' vanish i.e. that
(b - a) sin 2θ + h cos 2θ = 0
or
Thus we must choose an angle of rotation given by 1) to eliminate the term in xy.
Simplification by translation of the coordinate system. The first degree terms can be eliminated from a second degree equation by a translation of the coordinate system providing the equation is that of a conic with a center. The conics with centers are the ellipses and hyperbolas. These are called central conics. If a conic has a center, translation of the coordinate system to that center will eliminate the terms in x and y. If we translate the coordinate system to the point (x0, y0) the equation
f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
becomes
2) ax' 2 + 2hx'y' + by' 2 + 2(ax0 + hy0 + g)x' + 2(hx0 + by0 + f)y' + f(x0, y0) = 0
in the new, translated coordinate system. Under what circumstance will the terms in x and y disappear under this translation? By inspection it can be seen that they will disappear for a point (x0, y0) that satisfies the system
3) ax0 + hy0 + g = 0
hx0 + by0 + f = 0
These are, in fact, the equations that give the center of a central conic. Now let us ask what the constant term c becomes on translation to a center. Let c' denote the constant term at the center. From 2) above we see that c' = f(x0, y0). What is the value of f(x0, y0) at a center? The function
f(x0, y0) = ax02 + 2hx0y0 + by02 + 2gx0 + 2fy0 + c
can be written as
4) f(x0, y0) = (ax0 + hy0 + g)x0 + (hx0 + by0 + f)y0 + (gx0 + fy0 + c)
and we can see from inspection of this equation and 3) above that the first two terms vanish at a center (x0, y0). Thus the new constant term c' has the value
c' = f(x0, y0) = gx0 + fy0 + c
Theorem. If
f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
is the equation of a central conic and a translation is made to its center (x0, y0), the new equation is
f(x', y') = ax' 2 + 2hx'y' + by' 2 + gx0 + fy0 + c .