Rewrite the equation in a rotated x'y'-system without an x'y' term. Can I get some help? Thanks!

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2 Answers
May 15, 2018

The second selection:

x^2/4+y^2/9=1

Explanation:

The given equation

31x^2+10sqrt3xy+21y^2-144=0" [1]"

is in the general Cartesian form for a conic section:

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

where A = 31, B = 10sqrt3, C = 21, D = 0, E = 0 and F = -144

The reference Rotation of Axes give us equations that allow us to rotate a conic section to a specified angle, theta. Also, it gives us an equation that allows us to force the coefficient of the xy to become 0.

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

Substituting the values from equation [1]:

theta = 1/2tan^-1((10sqrt3)/(21-31))

Simplify:

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

theta = -pi/6

Use equation (9.4.4b) to verify that new rotation causes the coefficient of the xy term to be 0:

B' = (A-C) sin(2theta) + B cos(2theta)

B' = (31-21) sin(2(-pi/6)) + 10sqrt3cos(2(-pi/6))

B' = 0 larr verified.

Use equation (9.4.4a) to compute A':

A' = (A + C)/2 + [(A - C)/2] cos(2theta) - B/2 sin(2theta)

A' = (31 + 21)/2 + [(31 - 21)/2] cos(2(-pi/6)) - (10sqrt3)/2 sin(2(-pi/6))

A' = 36

Use equation (9.4.4c) to compute C':

C' = (A + C)/2 + [(C - A)/2] cos(2theta) + B/2 sin(2theta)

C' = (31 + 21)/2 + [(21 - 31)/2] cos(2(-pi/6)) + (10sqrt3)/2 sin(2(-pi/6))

C' = 16

Use Equation (9.4.4f) to compute F'

F' = F

F' = -144

Now, we can write the unrotated form:

36x^2+16y^2-144 = 0

Divide both sides by 144:

x^2/4+y^2/9-1 = 0

Add 1 to both sides:

x^2/4+y^2/9=1

May 15, 2018

Option B

Explanation:

We can write the equation in matrix form and then spin it onto its principal axis.

Let:

bb x^T M bb x = [x,y] [(a,b),(b,c)] [(x),(y)] = Q

= (x,y) [(ax + b y),(bx + cy)]= Q

= ax^2 + 2b xy + cy^2= Q

implies a = 31, d = 5 sqrt3, c = 21, Q = 144

And so in matrix form:

bb x^T [(31, 5 sqrt3),(5 sqrt3, 21)] bb x = 144 qquad square

To rotate the axes bbx by theta:

bb x^' = R(theta) bb x

  • implies bbx = R^(-1) bbx^'

Transposing bb x^' = R bb x :

implies bb x^('^T) = ( R bbx)^T = bb x^T R^T

implies bb x^('^T) = bb x^T R^(-1), as R is orthogonal

  • implies bb x^('^T) R = bb x^T

Putting these last 2 results into square:

bb x^('^T) R [(31, 5 sqrt3),(5 sqrt3, 21)] R^(-1) bb x^' = 144

IOW if R is the matrix that diagonalises M , then we have the equation in terms of its principal axes for diagonal eigenvector matrix D, ie:

  • D = R M R^(-1)

M 's eigenvalues are 36 and 16 so it can be diagonalised as:

bb x^('^T) D bb x^' = bb x^('^T) [(36, 0),(0 , 16)] bb x^' = 144

(x', y') [(9, 0),(0 , 4)] ((x'),(y')) = 36

x^('^2)/4 + y^('^2)/9 = 1