Condition for second degree equation to be a cone

A cone with vertex (x₀, y₀, z₀) has an equation that is homogeneous of the second degree in (x − x₀), (y − y₀), (z − z₀). So we show that under the given condition the equation can be written in that form.

Complete the squares

Write the surface as

a x² + 2ux + b y² + 2vy + c z² + 2wz + d = 0
a(x² + 2ux/a) + b(y² + 2vy/b) + c(z² + 2wz/c) + d = 0
a(x + u/a)² − u²/a + b(y + v/b)² − v²/b + c(z + w/c)² − w²/c + d = 0

So

a(x + u/a)² + b(y + v/b)² + c(z + w/c)² + [ d − (u²/a + v²/b + w²/c) ] = 0

Apply the condition

If u²/a + v²/b + w²/c = d, the bracket vanishes and the equation becomes

a(x + u/a)² + b(y + v/b)² + c(z + w/c)² = 0

Shift the origin to the point V(−u/a, −v/b, −w/c), putting X = x + u/a, Y = y + v/b, Z = z + w/c:

aX² + bY² + cZ² = 0.

This is homogeneous of degree 2 in X, Y, Z. If (X, Y, Z) satisfies it, so does (kX, kY, kZ) for every k. So every point on the surface lies on a line through V that lies wholly on the surface: the surface is a cone with vertex V(−u/a, −v/b, −w/c). ∎