# Find the points on the surface y^{2} = 49 + xz that are closest to the origin.

**Solution:**

The distance between an arbitrary point on the surface and the origin is

d(x, y, z) = √x^{2} + y^{2} + z^{2}

Here we have to minimize x^{2} + y^{2} + z^{2} to y^{2} = 49 + xz, it's easy to show that √f(x) and f(x) share the same critical points.

We can consider augmented distance function, D(x,y,z) = √x^{2} + y^{2} + z^{2}

Using Lagrange multipliers,

L(x, y, z, λ) = x^{2} + y^{2} + z^{2 }+ λ(y^{2 }- 49 - xz)

We have partial derivatives

L_{x} = 2x - λz

L_{y} = 2y + 2yλ

L_{z} = 2z - λx

L_{λ} = y^{2 }- 49 - xz

Set each partial derivative to zero to find critical points.

L_{y} = 0

⇒ 2y + 2yλ = 0

y = 0 and λ = -1

Put λ = -1 in Lₓ and Lz

L_{x} = 0

⇒ 2x - λz = 0

⇒ 2x + z = 0 --- (1)

L_{z} = 0

⇒ 2z - λx = 0

⇒ 2z + x = 0 --- (2)

Solving (1) and (2)

We get, x = 0 and z = 0

Put the values of x and z in y^{2} = 49 + xz

This means y^{2} = 49

⇒ y = ±7

Therefore, the points on the surface closest to the origin are (0, ±7, 0).

## Find the points on the surface y^{2} = 49 + xz that are closest to the origin.

**Summary:**

The points on the surface y^{2} = 49 + xz that are closest to the origin are (0, ±7, 0).

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