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Quadratic Forms

by stunner5000pt
Tags: forms, quadratic
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stunner5000pt
#1
Apr7-06, 04:31 PM
P: 1,439
Write the equation in terms of new caraibles so that it is in standard position and identify the curve

[tex] 3x^2 - 4xy = 2 [/tex]

here a = 3, b = -4, c = 0 , [tex] d = \sqrt{(-4)^2+(3-0)^2} = 5 [/itex]

[tex] \cos\theta = \frac{a+c-d}{\sqrt{b^2 + (a+c-d)^2}} = \frac{-2}{2\sqrt{5}} [/tex]
[tex] \sin\theta = \frac{b}{\sqrt{b^2 + (a+c-d)^2}} = \frac{4}{\sqrt{20}} [/tex]

so [tex] P = \frac{1}{\sqrt{5}} \left(\begin{array}{cc} -1&-2 \\ 2&-1 \end{array}\right) [/tex]

from X = PY i get
[tex] x = \frac{-1}{\sqrt{5}} (x_{1}-2y_{1}) [/tex]
[tex] y = \frac{-1}{\sqrt{5}} (2x_{1}+y_{1}) [/tex]
where x1 and y1 are the new variables
is this fine??

is this how you get the change of variables??
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