Please help with coordinate axes rotation!

Question

The exact question was:

"Rotate the coordinate axes to eliminate the xy term.

2x² + 2(sqrt3)xy + 4y² - 8 = 0"

 

(sqrt= square root)

 

 

Answer 1

For a conic in form Ax^2 + Bxy +Cy^2 +Dx+Ey+F=0

 

Rotation about an angle 0.5invcot(A-C)/B  eliminates the xy term.

 

So in this case the angle is 0.5cot-1/sqrt3 =-pi/6...30 degrees clockwise

 

This is a formula you can memorize for GRE Math Subject test.  Its derivation can be found at http://en.wikipedia.org/wiki/Rotation_of_axes

Answer 2

In order to eliminate the xy term, we rotate the axes by an angle θ. We are given the equation:

2x² + 2√3xy + 4y² - 8 = 0

For a conic, the equation is:

Ax² + Bxy + Cy² + Dx + Ey + F = 0

When we compare that with the equation given, it means:

A = 2

B = 2√3

C = 4

There are no x or y terms, so D = 0 and E = 0

To eliminate the xy term, we use:

tan(2θ) = B / (A - C)

This formula tells us what angle θ to rotate the axes by.

We can now substitute our values into this formula:

tan(2θ) = 2√3 / (2 - 4)

= 2√3 / (-2)

= -√3

Now, we need an angle whose tangent is -√3. We know that:

tan(60°) = √3

So, for tangent to be negative, the angle can be:

120° because tan(120°) = -√3. Therefore:

2θ = 120°

θ = 60°

So, we rotate the axes by 60°. We can use the rotation of axes formulas to rewrite the old variables x and y in terms of the new rotated variables x′ and y′. The formulas are:

x = x'cos(θ) - y′sin(θ)

y = x′sin(θ) + y′cos(θ)

Since cos 60° = 1/2 and sin 60° = √3/2. We get:

x = (1/2)x′ - (√3/2)y′

y = (√3/2)x′ + (1/2)y′

Lets substitute these into the original equation:

2x² + 2√3xy + 4y² - 8 = 0

After simplifying this, the equation becomes:

5x′² + y′² - 8 = 0 or 5x′² + y′² = 8

which is the final equation after rotating the axes by 60°, and the xy term is eliminated.