Show that the differential equations of all cones which have their vertices at $(0,0,0)$ is $px+qy=z$ Question: Show that the differential equations of all cones which have their vertices at $(0,0,0)$ is $px+qy=z$, where $p=\dfrac{\partial z}{\partial x}, q=\dfrac{\partial z}{\partial y}$
Progress: We have the cone with vertex at origin as $ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$. 
 A: Every straight line through the origin, if has a common point with such a cone lies completely on the cone.
Hence if $(l,m,n)$ lies on a cone $f(x,y,z) =0$, i.e. $f(l,m,n)=0$, then $f(tl,tm,tn)=0$ holds. Thus in particular if $f$ is a homogeneous function of degree $k$ then the above property holds, because $f(tl,tm,tn)= (t^k).f(l,m,n) =(t^k).0 =0$.
To keep things simple we restrict to cones of second degree, i.e. of the form : $ax^2+by^2+cz^2 =0$, to which it is possible to transform a general homogeneous equation of degree $2$ of the form $f(x,y,z)$ using reduction of quadratic forms. Differentiating the above equation with respect to $x$ and $y$ partially treating $z$ as a function of two variables and denoting the respective partial derivatives by standard notations of $p$ and $q$, we get:
$2ax +2czp =0$ and $2by +2czq =0$. This implies $ax=-czp$ and $by=-czq$. Using these in the given equation of the cone we get :
$-cxzp -cyzq +cz^2 = 0$ or  $z = xp + yq$ as the partial differential equation of first order for the class of all second degree cones with vertex at the origin and whose axis is the $z$-axis and with principal axes along the $x$- and the $y$-axes.
