Find the coordinates of the sationary points on the curve x^2 -xy+y^2=12

To find stationary points, we need to find dy/dx and set it equal to 0. Here we must use implicit differentiation: d/dx(x²) + d/dx(-xy) + d/dx(y²) = d/dx(12). Hence 2x - x(dy/dx) - y + 2y(dy/dx) = 0. Factorising: (2y - x)dy/dx = y - 2x . Hence dy/dx = (y - 2x)/(2y - x). dy/dx = 0: (y - 2x)/(2y - x) = 0 hence y - 2x = 0, y = 2x. Substituting this into the equation of the curve: x² - x(2x) + (2x)² = 12 So x² - 2x² + 4x² = 12. Thus 3x² = 12, x² = 4 so x = 2 or -2. When x = 2, y = 4 and when x = -2, y = -4 since y = 2x at the stationary points. So the coordinates of the stationary points are (2,4) and (-2,-4).