Find the equations of tangent plane, normal line to surface

[italia]

I need help with the following problem:

Find equations of the tangent plane and normal line to the surface x=3y[sup:2gpat5pq]2[/sup:2gpat5pq]+3z[sup:2gpat5pq]2[/sup:2gpat5pq]-273 at the point (-6,5,8). Tangent Plane: (make the coeffiecient of x equal to 1) ____________ = . Normal lin: (-6,__________,___________) +t(1,__________,___________)

 

[Deleted member 4993]

I need help with the following problem:

Find equations of the tangent plane and normal line to the surface x=3y[sup:1lrlr2l9]2[/sup:1lrlr2l9]+3z[sup:1lrlr2l9]2[/sup:1lrlr2l9]-273 at the point (-6,5,8). Tangent Plane: (make the coeffiecient of x equal to 1) ____________ = . Normal lin: (-6,__________,___________) +t(1,__________,___________)

Please show us your work, and exactly where you are stuck - so that we know where to begin to help you.

Do you know - what is a gradient vector?

 

[italia]

i've figured out the tangent plane is x-30y-48z+540 = 0. But i'm not sure how to figure out the normal line

 

[galactus]

\(\displaystyle x-3y^{2}-3z^{2}=278\)

Take the partials and get:

\(\displaystyle i-6yj-6zk\)

At (-6,5,8) we have \(\displaystyle {\nabla}F(-6,5,8)=i-30j-48k\)

Symmetric equations.

\(\displaystyle \frac{x+6}{1}=\frac{y-5}{-30}=\frac{z-8}{-48}\)

You can change to a vector equation of a line.