Question

Find the points on the cone z2 = x2 + y2 that are closest to the point (6, 2, 0).

Answer 1

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)