Several submissions used linear algebra to rotate this conic section. This is a perfectly acceptable and elegant solution -- however, this problem asked for a solution using the definition of a hyperbola. Here's the lengthy, tedious solution:

We have a hyperbola centered at the point (3,1) - therefore, the rotated conic will be centered at (3,1), as well. Let's ignore this for now and translate the center of the hyperbola to the origin.

Note: in a hyperbola, the sum of the squares of the major and transverse axes is equal to the square of the distance along the major axis between the center and both foci. We have: c² = 9 + 16 = 25, so c = 5.

The foci, then, are (-5,0) and (5,0). Consider the segment as a diameter of a circle with endpoints at these two foci, and rotate this diameter 45 degrees counterclockwise to get the new values of the foci, (5/√2, 5/√2) and (-5/√2, -5/√2).

Now, we can derive the equation of this new conic section. Recall that the difference of the distances from every point on the hyperbola to the respective foci will always be constant - the length of the major axis, 2*3 = 6.

Therefore, √((x - 5/√2)² + (y - 5/√2)²) - √((x + 5/√2)² + (y + 5/√2)²) = 6

This equation can quickly be put into standard form by adding √((x + 5/√2)² + (y + 5/√2)²) to both sides, squaring both sides, subtracting like terms to isolate the radical, dividing both sides by two, again squaring both sides, and setting one side equal to zero. The result should be divisible by 2; dividing through by 2 gives our conic section centered at the origin. (Note that squaring both sides twice has already eliminated the case where √((x - 5/√2)² + (y - 5/√2)²) - √((x + 5/√2)² + (y + 5/√2)²) = -6 )

To recenter the conic at (3,1), simply replace all values of x with (x-3) and all values of y with (y-1). Simplifying leads us to

7x² + 50xy +7y² - 92x - 164y - 68 = 0