Find the equation of cone whose vertex is origin and whose generators pass through the section of the sphere $x^2+y^2+z^2+2x+2y+2z+5=0$ by the plane $x+y+z=1$?
I did in following ways but i am getting 2 different answers. I am not able to find the mistake
Method 1:
Let $\frac{x}{l}= \frac{y}{m}=\frac{z}{n} = r,$ a parameter, be generator of given cone. (lr,mr,mr) lies in both. This gives $(l^2+m^2+n^2)r^2+2(l+m+n)r+5 =0, lr+mr+nr=1 \implies 7(l+m+n)^2+l^2+m^2+n^2=0 \implies 4(l^2+m^2+n^2)+7(lm+ln+mn)=0$
Locus of l,m,n is required cone i.e., $4(x^2+y^2+z^2)+7(xy+xz+yz)=0$ --->(1)
Method 2:
Equation of surface thru given sphere and plane is $S+\lambda P = 0$ where S=0 is given sphere and P=0 is given plane.
THis combined surface pass thru orgin. this gives me $x^2+y^2+z^2+7x+7y+7z=0$ --->(2)
I know (2) does not represent cone equation. But i am not able to find out where i went wrong. Pls point me to my mistake
Setting aside the issue that the given equation doesn’t represent a real sphere, your second method can’t work. Adding a multiple of $P$ to $S$ only changes the linear and constant terms of $S$. This amounts to translating and scaling the sphere, so you will always end up with another sphere, not a cone.