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How do you rotate the axes to transform the equation #5x^2-sqrt3xy+4y^2=6# into a new equation with no xy term and then find the angle of rotation?

1 Answer
Feb 2, 2017


Here is a very helpful reference


Using the equation (9.4.1) from the reference, #Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0#, we observe that:

#A = 5, B = -sqrt(3), C = 4, D = 0, E=0, and F = -6#

The angle that the conic section is rotated is the opposite rotation of the angle that equation (9.4.6) calculates, because (9.4.6) tells you what angle you use to "un"rotate" the conic section. Therefore, the angle that you want is:

#theta = -1/2tan^-1(B/(C-A))#

#theta = -1/2tan^-1(sqrt(3)/(4-5))#

#theta = -1/2tan^-1(-sqrt(3))#

#theta = pi/6" radians or "30^@" " larr# This is angle that the conic section is rotated.

If you want to "un-rotate" the conic section, then you must use the negative of this, #theta = -pi/6" radians or "-30^@#

Equation (9.4.3) is the new equation that you want:

#A'(x')^2 + B'x'y' + C'(y')^2 + D'x' + E'y' + F' = 0.#

Here are the equations that you will need to write the new equation:

(9.4.2a) #x' = xcos(θ) - ysin(θ)#
(9.4.2b) #y' = xsin(θ) + ycos(θ)#

(9.4.4a) #A' = (A + C)/2 + [(A - C)/2] cos(2θ) - B/2 sin(2θ)#
(9.4.4c) #C' = (A + C)/2 + [(C - A)/2] cos(2θ) + B/2 sin(2θ)#
(9.4.4f) F' = F

Please notice that I left out, B, D, and E, because we know that they will be 0.