Elimination of xy-term from rotating axes

T

TonyC

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Aug 22, 2005
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I am having trouble rotating the axes to eliminate the xy-term.
8x^2+64xy+8y^2+12x+12y+9=0

I know Ax^2+Bxy+Cy^2+Dx+Ey+F=0
however, the professor is looking for an equation not an answer.

Here are my choices and I am stumped:
40x^2 + 12 sq rt2 x + 12 sq rt 2 y + 9 = 0

40x^2 - 24y^2 +12 sq rt2y +9 = 0

-24y^2 + 12 sq rt2x + 12 sq rt 2y + 9 = 0

40x^2 - 24y^ + 12 sq rt2x + 9 = 0

I am not sure what the process is to get to this point. I can run the basic Rotation Theorem for Conics and answer the problem with an angle degree.

PLEASE HELP!
:?
 
Why don't you run the basic rotation theorem for conics. If the problem statement is desibned correctly, the answer will be one of your choices. Why are you hesitating?

Let's see, if 'y' is the angle of rotation, then tan(2*y) = (2*64)/(8-8) !! Well, OK, that can be a little alarming. Don't let is scare you. Where does the tangent blow up? How about pi/2? If 2*y = pi/2, the rotation angle is y = pi/4. Guess what? sin(pi/4) = cos(pi/4) = sqrt(2)/2 and there are those square roots you were worried about.
 
Right On! Where were you struggling? It wasn't that evil A = C, was it?
 
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