Question

find the equation of the tangent plane and normal line to the surface z²=4(1+x²+y²)at( 2,2,6)​

Answer 1

Answer:

We have,

z²=4(1+x²+y²)

4(1+x²+y²)-z² =0

df/dx = 8x, df/dy = 8y, df/dz = - 2z

At point (2,2,6)

df/dx = 16, df/dy = 16, df/dz = -12

Equation of tangent plane

16(x−2)+16(y−2)-12(z-6)=0

Equation of normal is

(x-2)/16 = (y-2)/16 = -(z-6)/12