Question

How do you rotate the axes to transform the equation `2x^2+sqrt3xy-y^2=-10` into a new equation with no xy term and then find the angle of rotation?

Answer 1

Please read the reference and the explanation.

Explanation:

The general Cartesian form for a conic section is:

`Ax^2+Bxy+Cy^2+Dx+Ey+F=0" [9.4.1]"`

In the given equation, `A = 2, B = sqrt3, C =-1, D = 0, E =0, and F = 10`

The angle to rotate the conic section back into alignment with the axis is:

`theta = 1/2tan^-1(B/(C-A))" [9.4.6]"`

`theta= 1/2tan^-1(sqrt3/(-1-2))`

`theta = -pi/12"rad or " -15^@ larr` angle of rotation

Let `A' =`the un-rotated coefficient

`A' = (A + C)/2 + [(A - C)/2] cos(2θ) - B/2 sin(2θ)" [9.4.4a]"`

Substituting in values:

`A' = (2 + (-1))/2 + [(2 - (-1))/2] cos(-pi/6) - sqrt3/2 sin(-pi/6)`

`A'=1/2+3/2sqrt3/2+sqrt3/2 1/2`

`A' = 1/2+sqrt3`

We know that B' will be 0. (I checked it on a scratchpad)

`C' = (A + C)/2 + [(C - A)/2] cos(2θ) + B/2 sin(2θ)" [9.4.4c]"`

Substituting in values:

`C' = (2 + (-1))/2 + [((-1) - 2)/2] cos(-pi/6) + sqrt3/2 sin(-pi/6)" [9.4.4c]"`

`C' = 1/2 - 3/2sqrt3/2 - sqrt3/2 1/2`

`C' = 1/2-sqrt3`

`F' = F"`

The new equation is:

`(1/2+sqrt3)x^2+(1/2-sqrt3)y^2+10=0`

In the standard form of the equation of a hyperbola:

`y^2/(sqrt(10/(sqrt3-1/2)))^2-x^2/(sqrt(10/(sqrt3+1/2)))^2=1`