How do you rotate the axes to transform the equation x^2-4xy+y^2=1 into a new equation with no xy term and then find the angle of rotation?

2 Answers
Jan 24, 2017
  • -X^2 + 3Y^2 = 1

  • rotation is pi/4

Explanation:

We can write the original conic section in matrix form as:

(x,y) ((1, -2),(-2 ,1)) ((x),(y)) = 1

Or:

(x,y) A ((x),(y)) = 1 qquad triangle

If we could write this is the following form with a diagonal matrix Lambda using new rotated axes X and Y we would have no XY term, ie:

(X,Y) color(blue)(((a_(11), 0),(0 ,a_(22))) ((X),(Y))) = (X,Y) color(blue)(Lambda) ((X),(Y)) = 1

That is what we are going to do.

Firstly, for a clockwise rotation theta from orthogonal axes (x,y) to new orthogonal axes (X,Y), we know that:

((x),(y)) = ((cos theta, -sin theta),(sin theta ,cos theta)) ((X),(Y)) = R((X),(Y)) qquad square

Now, because R is orthogonal, R^T = R^(-1); and because the transpose of the product of some matrices is the product of their transposes in reverse order, so we say wrt square that:

(x, y) = (X,Y) R^(-1) qquad circ

We can now re-write triangle using square and circ as:

(X,Y) color(red)(R^(-1) A R)((X),(Y)) = 1

If R is the matrix that diagonalises A, ie if Lambda = R^(-1) A R, then we have achieved what we set out to do :) because:

(X,Y) color(red)(R^(-1) A R)((X),(Y)) = 1 implies (X,Y) color(red)(Lambda)((X),(Y)) = 1

As it happens, matrix A is diagonalisable:

  • lambda_1 = -1, mathbf alpha_1 = ((1),(1))

  • lambda_2 = 3, mathbf alpha_2 = ((-1),(1))

The diagonal matrix is Lambda = ((-1, 0),(0, 3)), and the matrix of eigenvectors is R = ((1, -1),(1, 1)) with R^(-1) = 1/2((1, 1),(-1, 1))

So:

(X, Y) ((-1, 0),(0, 3)) ((X),(Y)) = 1

or:

-X^2 + 3Y^2 = 1, which is a hyperbola based symmetrically about the Origin.

In terms of the rotation, we need to ensure that our R is orthogonal:

Lambda = R^(-1) A R = 1/2((1, 1),(-1, 1)) ((1, -2),(-2 ,1)) ((1, -1),(1, 1))

= ((1/sqrt2, 1/sqrt2),(-1/sqrt2, 1/sqrt2)) A ((1/sqrt2, -1/sqrt2),(1/sqrt2, 1/sqrt2))

= ((cos (pi/4), sin (pi/4)),(-sin (pi/4), cos (pi/4))) A ((cos (pi/4), -sin (pi/4)),(sin (pi/4), cos (pi/4)))

implies R = ((cos (pi/4), -sin (pi/4)),(sin (pi/4), cos (pi/4)))

So the rotation is pi/4

Jan 24, 2017

The angle of rotation =pi/4

Explanation:

The equations of transformations are

x=Xcostheta-Ysintheta

y=Xsintheta+Ycostheta

Therefore,

x^2+y^2-4xy=1

(Xcostheta-Ysintheta)^2+(Xsintheta+Ycostheta)^2-4(Xcostheta-Ysintheta)(Xsintheta+Ycostheta)=1

X^2cos^2theta+Y^2sin^2theta-cancel(2XYsinthetacostheta)+X^2sin^2theta+Y^2cos^2theta+cancel(2XYsinthetacostheta)-4(X^2sinthetacostheta+Y^2sinthetacostheta+XYcos^2theta-XYsin^2theta)=1

X^2+Y^2-4(X^2+Y^2)sinthetacostheta-4XY(cos^2theta-sin^2theta)=1

We eliminate the XY term

cos^2theta-sin^2theta=0

costheta=sintheta

so,

theta=pi/4

Therefore,

x=X/sqrt2-Y/sqrt2

y=X/sqrt2+Y/sqrt2

So,

(X/sqrt2-Y/sqrt2)^2+(X/sqrt2+Y/sqrt2)^2-4(X/sqrt2-Y/sqrt2)(X/sqrt2+Y/sqrt2)=1

X^2+Y^2-4(X^2/2-Y^2/2)=1

X^2+Y^2-2X^2+2Y^2=1

3Y^2-X^2=1

graph{(x^2+y^2-4xy-1)(3y^2-x^2-1)=0 [-10, 10, -5, 5]}