To find the gradient vector ∇g(x, y), we need to take the partial derivatives of g with respect to x and y, and then evaluate them at the point (2, 4):
∂g/∂x = 2xy^2 - 6
∂g/∂y = 2x^2y
∇g(x, y) = [2xy^2 - 6, 2x^2y]
So, at the point (2, 4), we have:
∇g(2, 4) = [2(2)(4)^2 - 6, 2(2)^2(4)] = [62, 16]
The tangent line to the level curve g(x, y) = 8 at the point (2, 4) is perpendicular to the gradient vector ∇g(2, 4), so we can use the point-normal form of the equation of a line to write the equation of the tangent line:
(x, y) = (2, 4) + t[62, 16]
where t is a parameter. To find the value of t that corresponds to the point on the line where g(x, y) = 8, we substitute the coordinates of this point into the equation of the line:
8 = (2 + 62t)^2 (4 + 16t)^2 - 6(2 + 62t)
Expanding this expression and simplifying, we get a quadratic equation in t:
1024t^4 + 24864t^3 + 186384t^2 + 482280t - 191/3 = 0
Using a numerical method or a graphing calculator to solve this equation, we find that t ≈ -0.093 or t ≈ -0.660. Therefore, the two points on the tangent line where g(x, y) = 8 are:
(2 + 62(-0.093), 4 + 16(-0.093)) ≈ (-4.78, 1.48)
(2 + 62(-0.660), 4 + 16(-0.660)) ≈ (-37.32, -4.56)
So, the equation of the tangent line to the level curve g(x, y) = 8 at the point (2, 4) is approximately:
(x, y) = (-4.78, 1