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