Page 333 - A Course in Linear Algebra with Applications

9.2: Quadratic Forms 317

It was shown in Example 9.1.1 that the eigenvalues of A are
3 and -1 and that A is diagonalized by the orthogonal matrix

S =
V2

T
Put X — S X where X has entries x' and y'; then X — SX
and we read off that

1
( '
X = {X y')
T2
y ^ (*- + •)

So here 9 = TT/4 and the correct rotation of axes for this conic
is through angle n/4 in an anticlockwise direction. Substitut-
ing for x and y in the equation of the conic, we get

3x' 2 - y' 2 + 2^/2x' - y/2y' - 1 = 0.

From this we can already see that the conic is a hyperbola.
To obtain the standard form, complete the square in x' and

y':
2
3(z' + ^ ) - ( y ' + y/2 2 6"
4;) }
'
3
'
™
Hence the equation of the hyperbola in standard form is
3x" 2 - " 2 = 7/6,
y
where x" = x' + ^2/3 and y" = y' + l/y/2. This is a hy-
perbola whose center is at the point where x' = — \/2/3 and
y' = —\j\J1\ thus the xy - coordinates of the center of the
hyperbola are (1/6, —5/6). The axes of the hyperbola are the
lines x" = 0 and y" = 0, that is, x + y = —2/3 and x — y = l.