Expert: Paul Klarreich - 2/27/2011

Question
Rotate the axes to eliminate the “xy – “ term of the equation: 6x2 + 60xy + 6y2 + 9x + 9y + 52 = 0
*the answer is 36x^2 - 24y^2 + 9(sqrt 2)y + 52 = 0  

I have done everything up to finding the x and y.  I put the new equations for the x's and y's into the original equation but I don't know what to do from here.

Answer
Questioner: Kalli
Country: United States
Category: Calculus
Private: No
Subject: Precalculus- Rotation of Axes
Question: Rotate the axes to eliminate the “xy – “ term of the equation: 6x2 + 60xy + 6y2 + 9x + 9y + 52 = 0
*the answer is 36x^2 - 24y^2 + 9(sqrt 2)y + 52 = 0  

I have done everything up to finding the x and y.  I put the new equations for the x's and y's into the original equation but I don't know what to do from here.
..............................
Your basic rotation angle, t, {THETA?} is defined by:
         A - C
cot 2t = -------
           B

and you have A = 6, B = 60, C = 6, so

cot 2t = 0

==>  2t = pi/2,  t = pi/4, or 45 deg.

Now the basic equations are:

x = X cos t - Y sin t

y = X sin t + Y cos t

Now cos t and sin t are both sqrt(2)/2   {write s2/2 }

x = s2/2(X - Y)

Y = s2/2(X + Y)

Now plug that into your equation:

6x2 + 60xy + 6y2 + 9x + 9y + 52 = 0

and simplify.

{Make sure to use a wide screen for this.}

6(1/2)(X - Y)2 + 60(1/2)(X - Y)(X + Y) + 6(1/2)(X + Y)2 + 9(s2/2)(X - Y + X + Y) + 52 = 0

3(X2 - 2XY + Y2) + 30(X2 - Y2) + 3(X2 + 2XY + Y2) + 9(s2/2)(2X) + 52 = 0

3X2 - 6XY + 3Y2 + 30X2 - 30Y2 + 3X2 + 6XY + 3Y2 + 9X(s2) + 52 = 0

3X2  + 3Y2 + 30X2 - 30Y2 + 3X2  + 3Y2 + 9X(s2) + 52 = 0

36X2 - 24Y2 + 9X(s2) + 52 = 0

Well, almost.  Perhaps I got something backwards.  I leave it to you to find it.