Answer:
The closest points on the cone are;
(6, 2, -√10) and (6, 2, √10)
Step-by-step explanation:
Let B(x, y, z) denote a point on the cone.
Therefore, the distance between the points (6, 2, 0) and B(x, y, z) is;
d = √[(x - 6)² + (y - 2)² + (z - 0)²]
d = √[(x - 6)² + (y - 2)² + z²]
Since we are given that z² = x² + y², we now have;
d = √[(x - 6)² + (y - 2)² + x² + y²]
Taking the square of both sides gives;
d² = [(x - 6)² + (y - 2)² + x² + y²]
x² is an increasing function. Thus, minimizing d is also the same as to minimize f (x, y) = d²
Thus, f' = 0. So;
df/dx = 2(x - 6) + 2x = 0
2x - 12 + 2x = 0
4x = 12
x = 12/4
x = 3
Similarly,
df/dy = 2(y - 2) + 2y = 0
2y - 4 + 2y = 0
4y - 4 = 0
4y = 4
y = 4/4
y = 1
Now,from earlier;
z² = x² + y²
Thus;
z = ±√(3² + 1²)
z = ±√10
Thus, the closest points on the cone are;
(6, 2, -√10) and (6, 2, √10)
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