"Consider the surface given by 4x^2-y^2+5z^2-10z=55.
a) Find all points on the surface at which the tangent plane is parallel to the plane 8x+y+15z=1.
b) Pick one of these points and give the equation of the tangent plane to the surface at that point."
Could someone help me with this? Thank you.
To find the points on the surface where the tangent plane is parallel to the given plane, we need to find the normal vector to the surface and then determine the conditions for it to be parallel to the normal vector of the given plane.
a) To find the normal vector to the surface, we'll first rewrite the equation of the surface in a standard form. Then we can find the partial derivatives and use them to determine the normal vector.
Given: 4x^2 - y^2 + 5z^2 - 10z = 55
To rewrite in standard form, move all terms to one side:
4x^2 - y^2 + 5z^2 - 10z - 55 = 0
Now, take the partial derivatives of this equation with respect to x, y, and z:
∂f/∂x = 8x
∂f/∂y = -2y
∂f/∂z = 10z - 10
The coefficients of x, y, and z in the partial derivatives give us the coefficients of the normal vector, N = (A, B, C):
A = 8
B = -2
C = 10
Similarly, the normal vector of the given plane is P = (8, 1, 15).
For two vectors to be parallel, their directions must be proportional. Hence, A/B = 8/-2 = -4 and B/C = -2/10 = -1/5.
Plugging these ratios into the equation A/B = B/C, we find:
-4 = -1/5
This equation is not true, which means there are no points on the surface at which the tangent plane is parallel to the given plane. Therefore, the answer to part a) is that there are no such points.
b) Since there are no points on the surface satisfying the condition, we cannot provide an equation for a tangent plane at any point on the surface.