1

Show that general equation of the cone which touches three coordinate planes is $$ \sqrt{fx} \pm \sqrt{gy} \pm \sqrt{hz} = 0$$

The solution I am looking at begins by saying that general equation of cone through coordinate axes is $$ fyz + gzx + hxy= 0 $$

It then proceeds to calculate the reciprocal cone of above to arrive at the result in the question. Why is the reciprocal required here? Wouldn't an equation through the coordinate axes also touch the planes?

Pedro
  • 125,149
  • 19
  • 236
  • 403
s_a94248
  • 171
  • 10
  • Do you mean you consider this question to which I was asking clarifications ? – Jean Marie Sep 12 '23 at 07:48
  • "Touching the three coordinate planes" isn't the same as "passing through the coordinate axes" (as you say in your last sentence). – Jean Marie Sep 12 '23 at 07:50
  • 1
    @JeanMarie Yes I was referring to the same question. I am still not clear about how the two are reciprocals of each other when reciprocal means "locus of normals through the vertex of cone to the tangent planes of the given cone." Also, cannot help be amused that this same question bugged you and another person six years ago x)) – s_a94248 Sep 12 '23 at 08:50

1 Answers1

5

(we will assume thereafter that $a,b,c$ are all nonzero).

There is no need to consider the reciprocal cone for this issue.

In fact, the general equation of such a cone is obtained by taking the product of the 4 expressions given to you :

$$\underbrace{(\sqrt{ax}+\sqrt{by}+\sqrt{cz})}_{f_0(x,y,z)}\underbrace{\color{red}{(\sqrt{ax}-\sqrt{by}+\sqrt{cz})}}_{f_1(x,y,z)}\underbrace{\color{green}{(\sqrt{ax}+\sqrt{by}-\sqrt{cz})}}_{f_2(x,y,z)}\underbrace{\color{blue}{(\sqrt{ax}-\sqrt{by}-\sqrt{cz})}}_{f_3(x,y,z)}=0\tag{1}$$

giving

$$f(x,y,z)=a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} - 2 \, a b x y - 2 \, a c x z - 2 \, b c y z=0 \tag{2}$$

In this way, the locus of points verifying $f(x,y,z)=0$ is the union of the locii of points verifying $f_k(x,y,z)=0, k=0,1,2,3.$

The three parts of the cone defined by the three equations $f_k(x,y,z)=0, k=0,1,2,3$ have a nice pictorial representation as one can see on the figure below with the corresponding colors red, green blue (a half of the cone only has been represented for the sake of simplicity). Please note that equation $f_0(x,y,z)=0$ doesn't play any role because it is verified by a single point, the apex of the cone $(x,y,z)=(0,0,0)\cdots$

enter image description here

The 3 different "generating" lines (separating the different colored regions), have the following resp. equations:

$$D_x: \ \ \begin{cases}cz&=&by\\x&=&0\end{cases} \ \ D_y: \ \ \begin{cases}cz&=&ax\\y&=&0\end{cases} \ \ D_z: \ \ \begin{cases}by&=&ax\\z&=&0\end{cases}$$

Let us check for example that $D_x$ is tangent to coordinate plane $x=0$ is tangent to the cone along $D_x$ :

The generic equation of the tangent plane in point $(x_0,y_0,z_0)$ is

$$a^{2} xx_0 + b^{2} yy_0 + c^{2} zz_0- \, a b (x y_0+x_0y) - \, a c (x z_0+x_0z) - \, b c (y z_0+y_0z)=0 \tag{3}$$

Setting for example $(x_0,y_0,z_0)=(0,c,b)$ (one of the points of $D_x$) in (3), we get the equation of coordinate plane $x=0$, as awaited.

Edit : it is interesting to note that one can "logicaly" proceed from (2) to (1) by first writing (2) in the following form :

$$(ax-by-cz)^2=4bcyz$$

from which :

$$ax-by-cz=\pm 2 \sqrt{by}\sqrt{cz}$$

$$(\sqrt{ax})^2=(\sqrt{by})^2+(\sqrt{cz})^2\pm 2 \sqrt{by}\sqrt{cz}$$

$$(\sqrt{ax})^2=\left(\sqrt{by}\pm\sqrt{cz}\right)^2$$

$$\sqrt{ax}=\mp\left(\sqrt{by}\pm\sqrt{cz}\right)$$

giving the different factors of (1).

Jean Marie
  • 88,997
  • Interesting connection with (barycentric coordinates description) of conics inscribed in a triangle here with a figure that can be found there. – Jean Marie Oct 07 '23 at 07:10
  • Connected $\left|\matrix{0&x^2&y^2&c^2\x^2&0&z^2&b^2\y^2&z^2&0&a^2\c^2&b^2&a^2&0}\right|=a^2x^2+b^2y^2+c^2z^2-2abxy-2acxz-2bcyz$ factorizes into $(ax + by + cz)(ax + by - cz)(ax - by + cz)(ax - by - cz)$ – Jean Marie Oct 07 '23 at 19:45