Find an equation of the tangent plane to the surface at the given point. \[ f(x, y)=x^{2}-2 x y+y^{2}, \quad(8,9,1) \]

To find the equation of the tangent plane to the surface given by \( f(x, y) = x^2 - 2xy + y^2 \) at the point \( (8, 9) \), we first need to compute the partial derivatives \( f_x \) and \( f_y \). 1. **Calculate Partial Derivatives:** - \( f_x = \frac{\partial}{\partial x}(x^2 - 2xy + y^2) = 2x - 2y \) - \( f_y = \frac{\partial}{\partial y}(x^2 - 2xy + y^2) = -2x + 2y \) 2. **Evaluate at the Point (8, 9):** - \( f_x(8, 9) = 2(8) - 2(9) = 16 - 18 = -2 \) - \( f_y(8, 9) = -2(8) + 2(9) = -16 + 18 = 2 \) 3. **Point on the Surface:** - Verify \( f(8, 9) \) to ensure it matches the z-value in the point (8, 9, 1): - \( f(8, 9) = 8^2 - 2(8)(9) + 9^2 = 64 - 144 + 81 = 1 \) (This checks out!) 4. **Equation of the Tangent Plane:** The equation of the tangent plane is given by: \[ z - f(a, b) = f_x(a, b)(x - a) + f_y(a, b)(y - b) \] Substituting \( a = 8 \), \( b = 9 \), \( f(8, 9) = 1 \), \( f_x(8, 9) = -2 \), and \( f_y(8, 9) = 2 \): \[ z - 1 = -2(x - 8) + 2(y - 9) \] Rearranging gives: \[ z = -2x + 2y + 1 \] Thus, the equation of the tangent plane at the point \( (8, 9, 1) \) is: \[ z = -2x + 2y + 1 \]