Which Points on 4x^2+5y^2+5z^2=1 Have Tangent Planes Parallel to 4x-37-1z=-2?

[DieCommie]

"Find the points on the surface 4x^2+5y^2+5z^2=1 at which the tangent plane is parallel to the plane 4x-37-1z=-2"

Im very lost when it comes to this problem. I know that if the planes are parallel, the normal vectors will be parallel. So I think I need to multiply plane (4x-37-1z=-2)'s normal vector by some consistent?

Im not really sure what to do, nor do I understand the concept. Any help would be appreciated.

 

[HallsofIvy]

Back again? As I said in an earlier post, Let $F(x,y,z)= 4x^2+5y^2+5z^2$ so that you can think of the surface as a "level surface for F: F(x,y,z)= 1. Then the gradient of F, $\nabla F= 8x i+ 10y j+ 10z k$ is normal to the surface. You need to find (x, y, z) that not only satisfy $4x^2+5y^2+5z^2= 1$ but so that
8x i+ 10y j+ 10z k is a multiple of 4i+ 3j- k (I assume that when you typed "37" you meant "3y")- that is, 8x= 4a, 10y= 3a, 10z= -a for some (x,y,z) and a with (x,y,z) satisfying $4x^2+5y^2+5z^2= 1$. Solve 8x= 4a for x, 10y= 3a for y, 10z= -a for z and substitute into the equation of the surface to get an equation in a.

 

[DieCommie]

Awsome, great instructions.

I was doing nearly all of that, but I had a transcription error similar to my typo. Also I wasnt sure where to plug "a" into, but you simply plug it into the 8x=4a eqns etc. Really helped, thanks a lot

Back again?

actually i posted this first. I am going to look over that other thread after work