Answer to Question #149010 in Analytic Geometry for Sarita bartwal

Question

Show that the cone whose vertex is at the origin and which passes through the curve vif intersection of the sphere x^2+y^2+z^2=3p^2 and any plane which is at the distance‘p' from the origin has mutually perpendicular generator s.

Answer 1

From : Text Book Of 3-D Sphere, Cone And Cylinder

By A.K. Sharma (pg. 157)

Any plane at a distance 'p' from the origin is

$$lx+my+nz=p -------(1)$$

(Hence $l,m,n$ are direction cosines of the normal to the plane). The equation of the cone whose vertex is the origin and which passes through the intersection of the given sphere of the plane (1) is;

$x^2+y^2+z^2=3p^2=3(lx+my+nz)^2\\ \implies (1-3l^2)x^2+(1-3m^2)y^2+(1-3n^2)z^2-6lmxy-6mnyz-6nlxy=0----(2)$

With the equation of the cone generally defined as

$$f(x,y,z)=ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$$

Then form

$f(x,y,z)=(1-3l^2)x^2+(1-3m^2)y^2+(1-3n^2)z^2-6lmxy-6mnyz-6nlxy=0$

and we get that

$a=(1-3l^2);\\ b=(1-3m^2);\\ c=(1-3n^2)$

$\therefore3-3(l^2+m^2+n^2)=3-3,1=0$

Hence the cone has three mutually perpendicular generators.