(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