This is rather difficult to describe verbally. Basically, it is known that the eccentricity of a rectangular hyperbola is √2 so prior to rotation of the axes the foci are S(±√2a, 0), i.e. S(±2c, 0). This means that after rotation counter-clockwise by 45 degrees, the new foci are now located at a distance of 2c from the origin on the line y = x. A right-angled isoceles triangle with hyp = 2c, opp = adj & opp^2 + adj^2 = 4c^2 can be constructed. Solving that gives the coordinates of the new foci, which are S'(±√2c, ±√2c). Similarly, for the directrices work out their gradient after rotation and perpendicular distance from the origin.
If you have access to Cambridge 4U, there's a lovely explanation in the Conics chapter (and a very nice proof of the equation of the rectangular hyperbola using complex numbers).