Find an equation of the tangent plane to the parametric surface x=2rcos(theta) y=-4rsin(theta) z=r at the point (2sqrt(2),-4sqrt(2),2) when r=2, theta=pi/4
z=
"hint: your answer should be an expression of x and y"
taking some derivatives we get
Evaluating them at the point
Now if we take the cross product of the partials we will get a normal vector to the parameteric surfact at the point.
So the normal vector is
Now let (x,y,z) be any point in the plane. Using the other point we calculated before we can create a vector in the plane
Since the above vector lies in the plane and we have a normal vector to the plane the dot product of these two vectors must be zero.(Becuase they are perpendicular)
simplifying