Three mutually perpendicular tangent planes to a paraboloid which intersect at a given point

Question

Inspired by this problem, consider the paraboloid

$$ z = \dfrac{1}{4} (x^2 + y^2) $$

And the point $P(2, 3, -2)$. I want to find a set of three mutually perpendicular planes tangent to the paraboloid and passing through $P$.

How can I do that ? This is the question.

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Initial thoughts:

The paraboloid equation can be put in this matrix-vector format

$ r^T Q r + b^T r = 0 $

where $r = [x,y,z]^T $ , $ Q = \dfrac{1}{4} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} $ and $ b = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} $

The normal vector $N$ to the surface of the paraboloid is given by

$ N = 2 Q r + b $

A tangent plane at $r$ will have this normal $N$, and in addition we want to pass through $P$. Hence, we can write that

$ N^T (r - P) = 0 $

And this gives

$ (2 Q r + b)^T (r - P) = 0 $

Expanding gives

$ 2 r^T Q r + b^T r - 2 r^T Q P - b^T P = 0$

Since $ r^T Q r + b^T r = 0 $ , then the above simplifies to

$ - b^T r - 2 r^T Q P - b^T P = 0 $

i.e.

$ (2 Q P + b )^T r + b^T P = 0 \tag{1}$

And this is an equation of a plane. Thus the locus of all points $r$ on the paraboloid that will have a tangent plane passing through $P$ lies in this plane. The intersection of the of the plane and the paraboloid gives an ellipse.

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This ellipse of intersection can be described as follows

$ L(t) = C + V_1 \cos t + V_2 \sin t $

Since a possible tangency point is any point on this ellipse, let's pick a random value for $t$. Let this value be $t_1$. Then the first tangency point is

$ r_1 = C + V_1 \cos t_1 + V_2 \sin t_1 $

The normal vector at that point, which is also the normal vector to the first tangent plane is

$ n_1 = 2 Q r_1 + b $

Now we want to look for the other tangency points $-$ which below to the same ellipse, such that the normal vector at those points is perpendicular to $n_1$. So, if $r_2$ is the tangency point at $t = t_2$, then

$ r_2 = C + V_1 \cos t_2 + V_2 \sin t_2 $

And the normal vector at $r_2$ is

$ n_2 = 2 Q (C + V_1 \cos t_2 + V_2 \sin t_2 ) + b $

Since we want $n_2$ to be perpendicular to $n_1$, then the equation for $t_2$ is

$ n_1 \cdot \bigg( 2 Q (C + V_1 \cos t_2 + V_2 \sin t_2 ) + b \bigg) = 0 $

And this is a simple trigonometric equation in $t_2$ whose solutions are available in closed form, as a function of the coefficients of $\cos t_2$ and $\sin t_2$ and the constant term. It has two solutions which are the values of $t$ at the second and third tangency points.

Hence, we now have a complete set of normal vectors.

This shows that there is an infinite number of sets of three mutually perpendicular planes tangent to the given paraboloid and passing through the given point. Our selection of the first tangency point will determine the second and third tangency point from the above trigonometric equation.

The .GIF image below illustrates these mutually perpendicular planes to the given paraboloid, that pass through the given point.

Comments on the above discussion and/or alternative solutions are highly appreciated.

Answer 1

Following my comment, I have writen a Sage program where I have translated the paraboloid by vector $(2,3,-2)$ in order to place the apex of the tangent cone to the paraboloid at $(0,0,0)$ ; this process should give a cone to which each triple of "mutually orthogonal axes" (the intersection of mutually orthogonal tangent planes) should belong. See figure below with the program which has generated it. Beyond that, this program contains all the methodology in the commented lines (with a $\#$). Please note that I use $4 \times 4$ matrices.

It remains to find 3 orthogonal axes belonging to the cone...

Fig. 1 : Paraboloid, cone (in green) and plane (in blue) passing through their intersection.

Here is the SAGE program that can be executed on your own computer by calling https://sagecell.sagemath.org/ copying it the edit window and hitting the button "Evaluate" :

    var('x,y,z,x0,y0,z0,a,b,L,t')
    x0=2;y0=3;z0=-2;
    # (Shifted) Paraboloid's equation
    pa=expand((x+x0)^2+(y+y0)^2-4*(z+z0));show(pa)
    # pa=x^2 + y^2 + 4*x + 6*y - 4*z + 21;
    Mpa=matrix([[1,0,0,2],
                [0,1,0,3],
                [0,0,0,-2],
                [2,3,-2,21]]); # matrix of paraboloid
    # Cone's equation :
    cop=pa.subs(x=t*x,y=t*y,z=t*z);cc=cop.coefficients(t, sparse=0);
    co=expand((cc[1]/2)^2-cc[0]*cc[2])
    show(co)
    co=17*x^2 + 12*y^2 - 12*x*y + 8*x*z + 12*y*z - 4*z^2
    Mco=matrix([[17,-6,4,0],
                [-6,12,6,0],
                [4,6,-4,0],
                [0,0,0,0]]); # matrix of cone
    # Mpl=L*Mpa+Mco; 
    # We are looking for a degenerated quadric in this pencil
    s=solve(Mpl.determinant()==0,L);show(s) 
    # two solutions L=0 (trivial),L=-21 giving 
    pl=-21*pa+co;show(factor(pl))
    pl=2*x+3*y-2*z+21;
    # pl = 0 is the equation of the plane of the inters. ellipse
    a=-10;b=10;
    gr=implicit_plot3d(pa, (x,a,b), (y,a,b), (z,a,b),
                        plot_points=80, color='red',opacity=0.2,aspect_ratio=[1,1,1]);
    gr+=implicit_plot3d(co, (x,a,b), (y,a,b), (z,a,b),
                        plot_points=80, color='lightgreen');
    gr+=implicit_plot3d(pl, (x,a,b), (y,a,b), (z,a,b),
                        plot_points=80, color='lightblue', opacity=0.7)
    gr