Question

In which quadrant does the graph of the parabola x²+y²−2xy−8x−8y+32=0
lie?

Answer 1

The graph of the parabola is a unit circle centered at (4, 4).

We can rewrite the given equation as:

(x^2 - 2xy + y^2) - 8x - 8y + 32 = 0

Completing the square, we get:

(x - y)^2 - 8(x - y) + 4 = 0

Adding 16 to both sides, we have:

(x - y - 4)^2 = 12

Therefore, the graph of the parabola is a unit circle centered at (4, 4). Since the unit circle passes through (4, 4) and its reflection across the y-axis, the graph lies in the first and third quadrants.