find (if any) the equations of two planes tangent to the graph of f(x,y)=9-x^2-y^2 and perpendicular to each other

we can use the following formula: f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - (z - f(x_0, y_0)) = 0 where (x 0 ​ ,y 0 ​ ) is the point of tangency. The gradient of f is given by: ∇f = (-2x, -2y) Therefore, the equation of the tangent plane to the graph of f at the point (a,b) is: -2a(x - a) - 2b(y - b) - (z - (9 - a^2 - b^2)) = 0 For the two tangent planes to be perpendicular, the dot product of their normal vectors must be zero. The normal vector to the first tangent plane is (−2a,−2b,1), and the normal vector to the second tangent plane is (−2c,−2d,1). Therefore, we have the following equation: (-2a, -2b, 1) · (-2c, -2d, 1) = 0 This expands to: 4ac + 4bd + 1 = 0 We can now choose two different points (a,b) and (c,d) on the graph of f such that the above equation is satisfied. For example, we can choose (a,b)=(1,1) and (c,d)=(−1,−1). This gives us the following equations for the two tangent planes: -2(x - 1) - 2(y - 1) - (z - 7) = 0 2(x + 1) + 2(y + 1) - (z - 7) = 0 Simplifying these equations, we get: -2x - 2y - z + 9 = 0 2x + 2y - z + 9 = 0