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How to write an equation of cone formed by lines passing through the point $K(a,b,c)$ and intersecting the parabola $y^{2}=2px, z=0$ ?

WalrusYT
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  • It doesn't seem at first glance the data will uniquely specify a cone. Although lines of the cone pass through the given common point, $K(a,b,c)$, intersections with the parabola would seem restricted to the $x,y$ plane. – hardmath Jun 27 '22 at 18:09
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 27 '22 at 18:10
  • The vertex of the cone must be some $V=(-v,0,u)$, a generatrix is parallel to $x$ axis and the opposite generatrix passes through the origin. – Intelligenti pauca Jun 27 '22 at 21:41
  • Related: "What is the cone of the conic section?". In particular, see my answer, which covers the case of an ellipse; adapting the construction for a parabola is an interesting exercise. – Blue Jun 27 '22 at 23:30

1 Answers1

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The parametric equation of the parabolic cross section is

$ p(t) = ( 2 p t^2, 2 p t, 0 )$

lines connecting $K$ to $p(t)$ has the parametric equation

$ q(t,s) = K + s (p(t) - K) $

Now

$p(t) - K = ( 2 p t^2 - a, 2 pt - b , - c) \\ = (-a, -b, -c) + t (0, 2 p , 0) + t^2 (2p, 0, 0 ) $

Take the vector $[1, t, t^2]$. It lies on the quadric

$[1 , t , t^2] Q [ 1, t , t^2 ]^T = 0$ (Identically zero, $\forall t$)

Suppose $Q$ is given by

$Q =\begin{bmatrix} a && d && e\\ d && b && f \\ e && f && c \end{bmatrix} $

This results in

$ a + 2 d t + 2 e t^2 + b t^2 + 2 f t^3 + c t^4 = 0 $

which implies that $ a = d = f = c = 0 $ and $ b = - 2 e $.

So a possible $Q$ is

$Q =\begin{bmatrix} 0 && 0 && 1\\ 0 && -2 && 0 \\ 1 && 0 && 0 \end{bmatrix} $

So, now we have the following situation,

$p(t) - K = A u $

where $ A = \begin{bmatrix} -a && 0 && 2 p \\ -b && 2 p && 0 \\ -c && 0 && 0 \end{bmatrix} $

and $ u = \begin{bmatrix} 1 \\ t \\ t^2 \end{bmatrix}$

And as derived above, we have $ u^T Q u = 0 $ which implies that

$ \bigg( A^{-1} (p(t) - K) \bigg)^T Q \bigg( A^{-1} (p(k) - K) \bigg) = 0 $

Therefore, the equation of the desired cone is

$ ( q - K)^T Q_1 (q - K) = 0 $

where $ Q_1 = A^{-T} Q A^{-1} $

Hosam Hajeer
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