Find the points on the surface 3x2+5y2+3z2=13x2+5y2+3z2=1 at which the tangent plane is parallel to the plane 1x+1y+4z=71x+1y+4z=7.

The points are and

Step-by-step explanation:

We can find the normal to both the surface and the plane, and set a relation between them since tangent plane and the given are parallel. Once we have that we can use the original surface to find the points on the surface that are tangent to the plane.

Normal vectors.

The normal vectors can be found using partial derivatives or gradient, thus we can call the surface:

So its normal is

And the normal vector to the plane is

Notice that the normal vector of the plane is just represented by the coefficients of x,y and z.

Parallel condition

The two planes are parallel only if their normal vectors are parallel as well, so we can write

Using the previously found normal vectors.

So we get

Finding the points where we have the tangent plane.

We can plug the expressions we have found of x, y and z on the given  surface equation.

And we can simplify and solve for t.

The LCD is 60 so we can multiply all terms in both sides by 60 to get

So we get

That give us the values of t

Lastly at we have:

So one point is

And for the value of  

We have

So the second point is

omg what happend to your question? if you could edit this question i can

step-by-step explanation: