# Find the points on the cone z^{2} =x^{2} + y^{2} that are closest to the point (6, 2, 0)?

**Solution:**

Let us assume B(x, y, z) denotes a point on the cone.

So, 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} + z^{2}]

From the question it is given that z^{2} = x^{2} + y^{2},

We have,

d = √[(x - 6)^{2} + (y - 2)^{2} + x^{2} + y^{2}]

Taking the square of both sides, we get,

d^{2} = [(x - 6)^{2} + (y - 2)^{2} + x^{2} + y^{2}]

x^{2} is an increasing function.

Thus, minimizing d is also the same as to minimize f (x, y) = d^{2}

Thus,

f' = 0. So;

df/dx = 2(x - 6) + 2x = 0

2x - 12 + 2x = 0

4x = 12

x = 12/4

x = 3

Similarly follow same for finding y,

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

2y - 4 + 2y = 0

4y - 4 = 0

4y = 4

y = 4/4

y = 1

Then,

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

Thus;

z = ± √(3^{2} + 1^{2})

z = ± √10

Therefore, the points on the cone z^{2} =x^{2} + y^{2} that are closest to the point (6, 2, 0) are (6, 2, -√10) and (6, 2, √10).

## Find the points on the cone z^{2} =x^{2} + y^{2} that are closest to the point (6, 2, 0)?

**Summary:**

The points on the cone z^{2} =x^{2} + y^{2} that are closest to the point (6, 2,0) are (6, 2, -√10) and (6, 2, √10).

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