Given f(x,y,z)=x2−2y2+2z2−8=0 with point P(a,b,c) .
We calculate (fx,fy,fz)=(2x,−4y,4z) in this point: (2a,−4b,4c) .
Equation of the tangent plane is 2a(x−a)−4b(x−b)+4c(x−c)=0 , ax−2by+2cz=8.
The system {2x+3y+2z=8x−y+2z=5 is a line, we choose points on the line.
The points (3,0,1) and (−1,1,27) of the given line must lie in this tangent plane,
then we have c=28−3a and b=440−23a .
Using the equation a2−2b2+2c2−8=0 , we do not find a real solution for a .
Answer: The tangent plane does not exist.
Comments
Leave a comment