Where on a Hyperboloid Do Tangent Planes Align with a Given Plane?

[AndreTheGiant]

Homework Statement

Find the points on the hyperboloid 9x^2 -45y^2+5z^2 = 45 where the tangent plane is parallel to x+5y-2z = 7.

Homework Equations

The Attempt at a Solution

Ok so the tangent plane is parallel to the x+5y-2z=7 when their normal vectors are parallel. So that means that the normal vector for the given plane (1,5,-2) should also be a normal vector for the tangent plane to the hyperboloid.

So since the equation of the plane can be written as

(1,5,-2) . (x-x0,y-y0,z-z0) = x - x0 + 5y -5y0 -2z + 2z0 where -x0 - 5y0 +2z0 = -7

and the equation of the tangent plane to the hyperboloid can be written as the gradient at (x0,y0,z0) . (x-x0,y-y0,z-z0)

the gradient is (18x0, -90y0, 10z0) at (x0,y0,z0). and it has to be equal to the normal vector (1,5,-2).

So 18x0 = 1, -90y0 = 5, 10z0 = -2? and then solve for x0,y0,z0 and that is the point I'm looking for?

Just wanted to know if I'm doing this right.

 

[mighty2000]

Yes, you are on the right track with your approach. However, there are a few corrections and clarifications that I would like to make.

Firstly, when you say that the normal vector for the given plane should also be a normal vector for the tangent plane to the hyperboloid, this is not entirely correct. The normal vector for the given plane (1,5,-2) is a direction vector for the plane, but it is not a normal vector. To find the normal vector, we need to take the coefficients of x, y, and z in the plane equation and form a vector (1,5,-2), which is the normal vector.

Secondly, when you say that the equation of the plane can be written as (1,5,-2) . (x-x0,y-y0,z-z0) = x - x0 + 5y -5y0 -2z + 2z0, this is also not entirely correct. The equation of a plane in vector form is (x-x0,y-y0,z-z0) . n = 0, where n is the normal vector and (x0,y0,z0) is a point on the plane. So in this case, the equation of the given plane would be (x-x0,y-y0,z-z0) . (1,5,-2) = 0.

Now, to find the point on the hyperboloid where the tangent plane is parallel to the given plane, we need to find a point (x0,y0,z0) that satisfies both equations: 9x0^2 -45y0^2+5z0^2 = 45 and (x-x0,y-y0,z-z0) . (1,5,-2) = 0.

So, as you correctly identified, the gradient of the hyperboloid at (x0,y0,z0) should be parallel to the normal vector of the given plane (1,5,-2). This means that the gradient vector (18x0, -90y0, 10z0) should be a scalar multiple of (1,5,-2). This leads to the equations: 18x0 = k, -90y0 = 5k, 10z0 = -2k, where k is the scalar multiple.

Solving these equations, we get x0 = k/18