Graph of x^2 + 10xy + y^2 = 24

Here we outline the steps used in class on Monday, November 29.

> with(linalg):

> with(plots):

Find eigenvalues, eigenvectors of the matrix for the quadratic form.

> eigenvectors(matrix([[1,5],[5,1]]));

Because one eigenvalue is positive and the other negative we know the graph is a hyperbola. The diagonalization is [6 0 , 0 -4]. In the rotated coordinates we have 6x^2 - 4y^2 =24. One can check that the asymptotes are y = (sqrt(6)/2)x and y = -(sqrt(6)/2)x. A transition matrix is [1 1, 1 -1]. This is not quite a rotation matrix, but we can chose a [-1 1] as the eigenvector for -4 and get [1 -1, 1 1]. Now we get a rotation of 45 degrees. We normalize this by dividing through by sqrt(2). To get the unrotated asymptotes compute the slopes as follows.

> slope1:=evalf(tan(Pi/4+arctan(sqrt(6)/2)));

> slope2:=evalf(tan(Pi/4+arctan(-sqrt(6)/2)));

> M:=10; #range for plotting

> hyp:=implicitplot(x^2 +10*x*y + y^2 = 24,
x=-M..M,y=-M..M,color=black, thickness=3):

> a1:=plot(slope1*x,x=-M..M,yview=-M..M,
thickness=2,color=black,linestyle=3):

> a2:=plot(slope2*x,x=-M..M,yview=-M..M,
thickness=2,color=black,linestyle=3):

> display(hyp,a1,a2);

>