Cone with vertex (α,β,γ) and parabola guiding curve

Cone. Vertex V(α, β, γ); guiding curve the parabola y² = 4ax, z = 0.

Generator

A line through V with direction ratios (l, m, n):

(x − α)/l = (y − β)/m = (z − γ)/n. ... (1)

It meets the plane z = 0 where (z − γ)/n = −γ/n. So the point is

x₁ = α − γ l/n,   y₁ = β − γ m/n,   z = 0

This point is on the parabola: y₁² = 4a x₁.

(β − γ m/n)² = 4a (α − γ l/n)       ... (2)

Eliminate l, m, n

From (1), l/n = (x − α)/(z − γ) and m/n = (y − β)/(z − γ).

α − γ l/n = [α(z − γ) − γ(x − α)]/(z − γ) = (αz − γx)/(z − γ)
β − γ m/n = [β(z − γ) − γ(y − β)]/(z − γ) = (βz − γy)/(z − γ)

Put in (2):

(βz − γy)²/(z − γ)² = 4a (αz − γx)/(z − γ)

Answer: the cone is (βz − γy)² = 4a (αz − γx)(z − γ).

Check: at z = 0 this reduces to γ²y² = 4a γ² x, i.e. y² = 4ax.