Answer to Question #152620 in Calculus for iki

Question

4(x+1)^2 +2(y-2)^2 +(z-1)^2=1

Answer 1

4(x+1)² +2(y-2)² +(z-1)²=1

Finding the equation of the tangent plane to the surface which passes through the point (-1,2,2)

Equation of the tangent plane is

F_x(x_o,y_o,z_o)(x-x_o) +F_y(x_o,y_o,z_o)(y-y_o) + F_z(x_o,y_o,z_o)(z-z_o)

x_o=-1 ,y_o=2 ,z_o=2

F(x,y,z)=4(x+1)² +2(y-2)² +(z-1)²-1=0

F_x= 8x + 8 (i.e. partial derivative with respect to x)

F_x(-1,2,2)= 8(-1)+8=0

F_y= 4y - 8(i.e. partial derivative with respect to y)

F_y(-1,2,2)= 4(2) -8= 0

F_z= 2z - 2 (i.e. partial derivative with respect to z)

F_z(-1,2,2)= 2(2)-2=2

Equation of the tangent plane is

F_x(x_o,y_o,z_o)(x-x_o) +F_y(x_o,y_o,z_o)(y-y_o) + F_z(x_o,y_o,z_o)(z-z_o)=0

0(x+1) + 0(y-2) + 2(z-2)=0

2z - 4 =0

2z - 4 =0

The equation of the tangent that passes through the point (-1,2,2) is

2z - 4 =0