condition for cones to be reciprocal Question : Show that the cone $$ax^2 + by^2 + cz^2 - cxy - ayz - bzx = 0$$ is the reciprocal of the cone $$(a^2 - bc)x^2 + (b^2 - ac)y^2 + (c^2 - ab)z^2 - 2(a^2 + bc)yz - 2(b^2 + ac)zx - 2(c^2 + ab)xy =0$$
To me it seems like the that they can be proved using determinants but my instructor didn't covered this portion of the syllabus. Also my books didn't have any such thing or even minutely releated concept.
I was going through the previous year question papers and I saw this question. 
Kindly give me a hint to solve this. Is there any hint how to show this using determinant. I want to learn how to solve this. 
 A: It certainly seems, as Jyrki suggests, that the matrix of the reciprocal cone is the inverse of the original matrix, or a scalar multiple so that we do not have lots of fractions hanging around. YES! The matrix can be taken to be the Hessian matrix of second partial derivatives. Here are selections from two books that talk about this, including some exercises with answers
The simple proof: evidently the reciprocal cone is made up of normal vectors to the original. Our original cone is given by $x^T H x = 0,$ where the Hessian matrix $H$ is symmetric, indefinite, and has nonzero determinant, so it is nonsingular. 
The gradient of $x^T H x,$ written as a column vector, is $2Hx.$ The reciprocal cone is, therefore, the collection of all $2Hx$ where $x$ satisfies $x^T H x = 0.$ Drop the $2$ and make a new name, $y = Hx,$ so that $y$ is in the reciprocal cone. What is $y^T H^{-1} y?$
Well,
$$ \color{blue}{ y^T H^{-1} y = (Hx)^T H^{-1} Hx = x^T H H^{-1} H x = x^T H x = 0}.   $$



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