Show that the cones $ayz+bxz+cxy=0$ and $(ax)^{1/2}+(by)^{1/2}+(cz)^{1/2}=0$ are reciprocal.

Question

Show that the cones $$ayz+bxz+cxy=0 \quad \text{and} \quad (ax)^\frac{1}{2}+(by)^\frac{1}{2}+(cz)^\frac{1}{2}=0$$ are reciprocal, the formar contains the co-ordinate axes while the latter touch the coordinte planes.

Answer 1

At some point I found the definition of reciprocal cone in this book. I posted some relevant pages at condition for cones to be reciprocal Alright, I was able to paste the same images here. Good to know how to do that, as I cannot seem to find those pages any more in the online book previews.

given cone $ayz + bzx + c xy = 0,$ the actual reciprocal cone is $$ a^2 x^2 + b^2 y^2 + c^2 z^2 - 2bcyz - 2cazx - 2abxy = 0. $$

About the coordinate planes, what happens when $z =0?$ Then $$ a^2 x^2- 2abxy + b^2 y^2 = (ax-by)^2 = 0, $$ so that $$ ax = by. $$ This is a single line, therefore the cone is tangent to the plane $z=0.$ Another way to confirm tangency is to point out that a point on this line is, for some $t,$ given by $(bt, at, 0).$ Furthermore, for a quadratic form given by $v^t H v,$ the gradient is given ( as a column vector) by $2 H v.$ We then calculate $$ \left( \begin{array}{rrr} a^2 & -ab & -ca \\ -ab & b^2 & -bc \\ -ca & -bc & c^2 \end{array} \right) \left( \begin{array}{r} bt \\ at \\ 0 \end{array} \right) = \left( \begin{array}{r} 0 \\ 0 \\ -2abct \end{array} \right) $$

$$ \left( \begin{array}{rrr} 0& c & b \\ c & 0 & a \\ b & a & 0 \end{array} \right) \left( \begin{array}{rrr} a^2 & -ab & -ca \\ -ab & b^2 & -bc \\ -ca & -bc & c^2 \end{array} \right) = \left( \begin{array}{rrr} -2abc & 0 & 0 \\ 0 & -2abc & 0 \\ 0 & 0 & -2abc \end{array} \right) $$

This seems to be a pretty good practice/solutions manual for this material

Answer 2

.I able to 1st part but i can not understand 2nd part.

Answer 3

Alright, it turns out this was just carelessness about square roots. IF $$ \color{blue}{ax \geq 0, \; b y \geq 0, \; cz \geq 0 }$$ AND $$ \color{blue}{ \sqrt {cz} = \sqrt {ax} + \sqrt {by}}, $$ square once to get $$ cz = ax + by + 2 \sqrt {abxy}, $$ so $$ - ax - by + cz = 2 \sqrt {abxy}. $$ Squaring again does give $$ a^2 x^2 + b^2 y^2 + c^2 z^2 - 2byz -2ca zx + 2abxy = 2abxy, $$ $$ a^2 x^2 + b^2 y^2 + c^2 z^2 - 2byz -2ca zx - 2abxy = 0. $$ This is a subset of the cone in just one octant, and is not the entirety of cone points in that octant. We get three pieces when $$ \color{blue}{ax \geq 0, \; b y \geq 0, \; cz \geq 0 },$$ namely $$ \color{blue}{ \sqrt {cz} = \sqrt {ax} + \sqrt {by}}, $$ $$ \color{blue}{ \sqrt {ax} = \sqrt {by} + \sqrt {cz} }, $$ $$ \color{blue}{ \sqrt {by} = \sqrt {cz} + \sqrt {ax} }. $$

There is no sense to taking all $+$ signs, as in $$ \sqrt {ax} + \sqrt {by} + \sqrt {cz} = 0, $$ this can only be a single point!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The example $a=b=c=1,$ with $x,y,z \geq 0,$ and $\sqrt z = \sqrt x + \sqrt y$ gives the satisfactory $$ z = x + y + 2 \sqrt {x y}. $$ This is part of a cone. We can write it in polar/cylindrical coordinates as $0 \leq \theta \leq \frac{\pi}{2},$ $$ z = r \; \left( \cos \theta + \sin \theta + 2 \sqrt { \cos \theta \; \sin \theta} \right) $$ Other points of the same cone in the first octant $x,y,z \geq 0,$ include $\sqrt x = \sqrt y + \sqrt z$ and $\sqrt y = \sqrt z + \sqrt x.$