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

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

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Answer 1 · 2017-08-09T15:08:43

A rotation through $\frac{2 π}{6}$ $(2pi)/6$ radians, (or $60^{o}$ $60^o$ ) will remove the $x y$ $xy$ -term

The new translated equation is:

$X^{2} - Y = 0$ $X^2 - Y=0$

Which represents a paraobla

Explanation:

We have:

$x^{2} + 2 \sqrt{3} x y + 3 y^{2} + 2 \sqrt{3} x - 2 y = 0$ $x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0$

Comparing to the conic equation in standard form:

$A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ $Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0$

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:

enter image source here

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

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

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Explanation:

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What rotation is required to remove the xyxyxy-term from the conic x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0x2+2√3xy+3y2+2√3x−2y=0x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0?

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