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# Help with rotation formulas?

I don t understand how to solve this and i d really appreciate if you can help step by step. Thank you

Write so that the equation has no xy term.

3x^2 - 10xy +3y^2 - 32 = 0

### 2 Answers

- ?Lv 74 years agoFavorite Answer
We need to rotate θ degrees.

Rotation matrix:

[X] = [ cosθ −sinθ ] [ x ]

[Y] [ sinθ cosθ ] [ y ]

[x] = [ cosθ sinθ ] [ X ]

[y] [ −sinθ cosθ ] [ Y ]

x = X cosθ + Y sinθ

y = Y cosθ − X sinθ

Now if we graph hyperbola, we can see that it has transverse axis y = −x

So we need to rotate 45° to get rid of xy term and get horizontal transverse axis. However, it might not always be possible to determine the angle of rotation, so I'll just solve by substituting x and y with the values above, then setting coefficient of XY = 0.

3x² − 10xy +3y² − 32 = 0

3(X cosθ + Y sinθ)² − 10(X cosθ + Y sinθ)(Y cosθ − X sinθ) +3(Y cosθ − X sinθ)² − 32 = 0

3(X² cos²θ + 2XY sinθ cosθ + Y² sin²θ) − 10(XY cos²θ − X² sinθ cosθ + Y² sinθ cosθ − XY sin²θ) + 3(Y² cos²θ − 2XY sinθ cosθ + X² sin²θ) − 32 = 0

3X² cos²θ + 6XY sinθ cosθ + 3Y² sin²θ − 10XY cos²θ + 10X² sinθ cosθ − 10Y² sinθ cosθ + 10XY sin²θ + 3Y² cos²θ − 6XY sinθ cosθ + 3X² sin²θ − 32 = 0

(3 cos²θ + 3 sin²θ + 10 sinθ cosθ)X² + (3 sin²θ + 3 cos²θ − 10 sinθ cosθ)Y² + (6 sinθ cosθ − 10 cos²θ + 10 sin²θ − 6 sinθ cosθ)XY − 32 = 0

(3 + 5 sin(2θ))X² + (3 − 5 sin(2θ))Y² − 10 cos(2θ)XY − 32 = 0

To get rid of XY term, we need:

cos(2θ) = 0 ----> 2θ = 90° ----> θ = 45°

sin(2θ) = sin(90°) = 1

8X² − 2Y² − 32 = 0

Graphs before and after:

- Randy PLv 74 years ago
You gave the method in your subject line. It uses rotation formulas. You're going to find new coordinate axes (x', y') by rotating the old ones, such that this curve doesn't have any x, y terms.

x = x' cos(theta) - y' sin(theta)

y = x' sin(theta) + y' cos(theta)

Plug those in.

3x^2 - 10xy + 3y^2 - 32 = 0

3[ x'^2 cos^2(theta) + y'^2 sin^2(theta) - 2x'y' cos(theta) sin(theta)]

-10[ x' cos(theta) - y' sin(theta) ][ x' sin(theta) + y' cos(theta)]

+ 3[ x'^2 sin^2(theta) + y'^2 cos^2(theta) + 2x'y' cos(theta) sin(theta)]

- 32 = 0

That's a big mess, but the x'y' terms in the first and third terms have opposite signs and the same coefficient (3) so they cancel out. So the only place where x'y' terms can originate is the second term.

10[ x' cos(theta) - y' sin(theta) ][ x' sin(theta) + y' cos(theta)] = 10[ x'^2 cos(theta) sin(theta) + x'y' cos^2(theta) - x'y' sin^2(theta) - y'^2sin(theta) cos(theta) ]

= 10(x'^2 + y'^2) sin(theta) cos(theta) + 10x'y' [ cos^2(theta) - sin^2(theta) ]

And only that last term is an x'y' term. We want that to be 0.

So cos^2(theta) = sin^2(theta) = cos(2*theta) has to be 0. That tells you how much the angle theta is.