# Find the points on the cone z^{2} = x^{2} + y^{2} that are closest to the point (4,2,0).

**Solution:**

Consider B (x, y, z) to represent the point on the cone.

So the distance between (4, 2, 0) and B (x, y, z) is

d = √[(x - 4)² + (y - 2)² + (z - 0)²]

d = √[(x - 4)² + (y - 2)² + z²]

We know that

z² = x² + y²

d = √[(x - 4)² + (y - 2)² + x² + y²]

Squaring on both sides

d² = [(x - 4)² + (y - 2)² + x² + y²]

x² is an increasing function.

Minimizing d is similar to minimize f (x, y) = d²

f' = 0

df/dx = 2(x - 4) + 2x = 0

2x - 8 + 2x = 0

4x = 8

x = 8/4

x = 2

In the same way

df/dy = 2(y - 2) + 2y = 0

2y - 4 + 2y = 0

4y - 4 = 0

4y = 4

y = 4/4

y = 1

From z² = x² + y²

z² = 2² + 1²

z = ±√(2² + 1²)

z = ±√5

Hence, the closest points on the cone are (2, 1, -√5) and (2, 1, √5).

## Find the points on the cone z^{2} = x^{2} + y^{2} that are closest to the point (4,2,0).

**Summary:**

The points on the cone z^{2}=x^{2}+y^{2} that are closest to the point (4,2,0) are (2, 1, -√5) and (2, 1, √5).