Final answer:
To find the tangent line to the ellipsoid at (1,2,2), we calculate the gradient at that point, yielding the normal vector (8, 8, 4). The equation of the tangent line is 8x + 8y + 4z = 32, which is not listed in the provided options.
Step-by-step explanation:
To find the tangent line to the ellipsoid 4x²+2y²+z²=16 at the point (1,2,2), we need to use the gradient of the ellipsoid. The gradient vector gives us the direction coefficients for the normal to the surface at the given point. To find the gradient, we partially differentiate the equation of the ellipsoid with respect to x, y, and z.
The partial derivatives are:
- ∂/(∂x) (4x² + 2y² + z²) = 8x
- ∂/(∂y) (4x² + 2y² + z²) = 4y
- ∂/(∂z) (4x² + 2y² + z²) = 2z
At the point (1,2,2), these derivatives evaluate to 8, 8, and 4, respectively. So the normal vector to the ellipsoid at the point is (8, 8, 4).
We know that a line is tangent to a surface if it is perpendicular to the normal of the surface at the point of tangency. This implies that the tangent line must satisfy the equation:
8(x-1) + 8(y-2) + 4(z-2) =0
Expanding and simplifying, we get the tangent line equation:
8x + 8y + 4z = 32
Since none of the given options matches this equation, it appears there may have been a mistake in the listed options or in the question as it was presented. Moreover, none of the given options could be the equation of the tangent line at the point (1,2,2) for the given ellipsoid.