Question

What rotation is required to remove the `xy`-term from the conic `x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0`?

Answer 1

A rotation through `(2pi)/6` radians, (or `60^o`) will remove the `xy`-term

The new translated equation is:

`X^2 - Y=0`

Which represents a paraobla

Explanation:

We have:

`x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0`

Comparing to the conic equation in standard form:

`Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0`

We have:

`A=1; \ B=2sqrt(3); \ C=3; \ D=2sqrt(3); \ E=-2; \ F=0`

The rotation angle, `theta`, that we seek to translate from the `xy`-plane to the `XY`-plane (say) and remove the `xy`-term is given by:

`cot 2theta = (A-C)/B = (1-3)/(2sqrt(3) = -1/(sqrt(3)`
`:. 2theta = (2pi)/3 => theta = (2pi)/6`

We will need `sin theta`, and `cos theta`, so let us calculate those also:

`sin theta = (sqrt(3))/2; \ \ \ \ cos theta = 1/2`

The rotation matrix to transform from `xy`-coordinate axis into the `XY`-coordinate axis is therefore:

`( (x),(y) ) = ( (cos theta, -sin theta), (sin theta, cos theta) ) ( (X),(Y) )`
`" " = ( (1/2, -(sqrt(3))/2), ((sqrt(3))/2, 1/2) ) ( (x'),(y') )`

Leading to the transformation equations:

`x = 1/2 X-sqrt(3)/2 Y; \ \ \ \ y = sqrt(3)/2 X + 1/2 Y`

We can substitute these transformation equations into the original equation to get the new equation (which should be deficient in an `XY`-term)

`(1/2 X-sqrt(3)/2 Y)^2+2sqrt(3)(1/2 X-sqrt(3)/2 Y)(sqrt(3)/2 X + 1/2 Y)+3(sqrt(3)/2 X + 1/2 Y)^2+2sqrt(3)(1/2 X-sqrt(3)/2 Y)-2(sqrt(3)/2 X + 1/2 Y)=0`

We can remove the fractions to make the algebra slightly less tedious, by multiplying by `4`

`(X-sqrt(3)Y)^2 + 2sqrt(3)(X-sqrt(3)Y)(sqrt(3)X + Y) + 3(sqrt(3) X + Y)^2 + 4sqrt(3)(X-sqrt(3)Y) - 4(sqrt(3) X + Y) = 0`

And now we multiply out and collect terms:

`(X^2-2sqrt(3)XY+3Y^2) + 2sqrt(3)(sqrt(3)X^2+XY-3XY-sqrt(3)Y^2) + 3(3X^2+2sqrt(3)XY+Y^2) + 4sqrt(3)(X-sqrt(3)Y) - 4(sqrt(3) X + Y)=0`

`:. X^2-2sqrt(3)XY+3Y^2 + 6X^2-4sqrt(3)XY-6Y^2 + 9X^2 + 6sqrt(3)XY + 3Y^2 + 4sqrt(3)X - 12Y - 4sqrt(3)X - 4Y=0`

`:. 16X^2 - 16Y=0`

`:. X^2 - Y=0`

Which in the `XY`-plane we can easily identify as a parabola.

The graphs of the two equation (on the same image) as as follows:

And we can graphically confirm that the original equation is the same parabola rotated through `(2pi)/6` radians, (or `60^o`)