# 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 }+ z^{2}]

We know that

z^{2} = x^{2} + y^{2}

d = √[(x - 4)^{2} + (y - 2)^{2} + x^{2} + y^{2}]

Squaring on both sides

d^{2 }= [(x - 4)^{2} + (y - 2)^{2} + x^{2} + y^{2}]

x^{2} is an increasing function.

Minimizing d is similar to minimize f (x, y) = d^{2}

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 z^{2} = x^{2} + y^{2}

z^{2} = 2^{2} + 1^{2}

z = ±√(2^{2} + 1^{2})

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).