Let T be the matrix of rotation counterclockwise by 45 degrees. Then based on what this does to the basic unit vectors, (1,0) and (0,1) the matrix T has columns [1/sqrt(2), 1/sqrt(2)], [-1/sqrt(2), 1/sqrt(2)]. Then if g = 0 is the equation for the original ellipse, and f=0 is the equation for the rotated ellipse, we have g(x) = 0 if and only if f(T(x)) = 0, if and only if (T*f)(x) = 0, where T* is pulling back functions via T. Hence g = T*f, so f = (T*)^-1(g). But pulling back polynomials is determined by the pullback on linear, i.e. coordinate functions, and pulling back linear functions is just done by the transpose matrix. But since T is a rotation matrix, if T* is the transpose matrix, then T* = T^-1, so (T*)^-1 = T. So since in the dual space the function x has coordinates (1,0) and y has coordinates (0,1), they transform by x-->x/sqrt(2) + y/sqrt(2) and y--> -x/sqrt(2) + y/sqrt(2). Thus the equation x^2/a^2 +y^2/b^2 =1, transforms into the equation obtained from these substitutions, i.e. (x+y)^2/2a^2 + (-x+y)^2/2b^2 = 1, where (-x+y)^2 = (x-y)^2.