Question

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

Answer 1

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`

Answer 2

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`