Question

show that differential equations of all cones which have their vertex at the origin is px+qy=z​

Answer 1

Answer:

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Step-by-step explanation:

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)(l,m,n) lies on a cone f(x,y,z)=0f(x,y,z)=0, i.e. f(l,m,n)=0f(l,m,n)=0, then f(tl,tm,tn)=0f(tl,tm,tn)=0 holds. Thus in particular if ff is a homogeneous function of degree kk then the above property holds, because f(tl,tm,tn)=(tk).f(l,m,n)=(tk).0=0f(tl,tm,tn)=(tk).f(l,m,n)=(tk).0=0.

To keep things simple we restrict to cones of second degree, i.e. of the form : ax2+by2+cz2=0ax2+by2+cz2=0, to which it is possible to transform a general homogeneous equation of degree 22 of the form f(x,y,z)f(x,y,z) using reduction of quadratic forms. Differentiating the above equation with respect to xx and yy partially treating zz as a function of two variables and denoting the respective partial derivatives by standard notations of pp and qq, we get:

2ax+2czp=02ax+2czp=0 and 2by+2czq=02by+2czq=0. This implies ax=−czpax=−czp and by=−czqby=−czq. Using these in the given equation of the cone we get :

−cxzp−cyzq+cz2=0−cxzp−cyzq+cz2=0 or z=xp+yqz=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 zz-axis and with principal axes along the xx- and the yy-axes.