Exploring Tangent Planes and Vertical Curvature in Differential Geometry

[ParisSpart]

Consider the surface S defined as the graph of a function z = 2x ^ 2 - y ^ 2
i) find a basis of the tangent plane Tp surface S at the point p = (-1,2, -2)
ii) find a non-zero vector w in Tp with the property that the vertical curvature at point p in the direction of vector w is zero


for the i) question i found that the tangent plane has this form: Tp={(-1,2,-2)+t(1,0-4)+s(0,1,-4)}
but how i will find a basis of this tangent plane? maybe is <(1,0,4),(0,1,-4)> ?

and for the ii) how i will find this non zero vector?

 

[HallsofIvy]

What is the "vertical curvature" at a point?