(4A-3) We know from questions (4A-1) and (4A-2) that the gradients of the two tangents that define this plane are:
∂z/∂x = -1 ∂z/∂y = -2
The function that defines the plane which, when differentiated, gives these partial derivatives, must be of the form z = -x - 2y + c, where c is a constant.
At the point where we found the two tangents which define this plane, we know that
x = y = -0.5
So the value of z at this point is given by z = x2 + 2y2 = 0.75.
This enables us to find the value of c in the equation of the plane, z = -x - 2y + c
When we substitute in these values, we get: 0.75 = 0.5 + 1.0 +c
From which we deduce that c = -0.75
Therefore the equation of the tangent plane is:
z = -x - 2y - 0.75