Answer to Question #91517 in Analytic Geometry for Ra

Question

Examine which of the following conicoids are central and which are non-central.
Also detrrmine which of the central conicoids have centre at the origin.
1. x²+y²+z²+x+y+z=1
2.2x²+4xy+xz-x-3y+5z+3=0
3.x²-y²-z²+xy+4yz+x=0

Answer 1

To determine central conicoid:

A conicoid (S) given by equation

S=ax²+by²+cz²+2fyz+2gzx+2hxy+2ux+2vy+2wz+d=0

has point p(x,y,z) as its center if

ax+hy+gz+u=0

hx+by+fz+v=0

gx+fy+cz+w=0.................(o)

These system of equations must have a unique solution

hence

$\begin{vmatrix} a & h & g \\ h & b & f\\ g & f & c \end{vmatrix}=0 \, (p)$

1. Answer

x²+y²+z²+x+y+z=1

As per condition (o)

x+1/2=0

y+1/2=0

z+1/2=0

Using matrix determinant property;

$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{vmatrix}=1 \ne 0$

(non zero)

Hence ,they have unique solution.

To find center of conicoid,solve the system of equations;

x=-1/2;y=-1/2;z=-1/2

Center is (-1/2,-1/2,-1/2).The conicoid is central.

2. Answer

2x²+4xy+xz-x-3y+5z+3=0

The equations will be

2x+2y+z/2-1/2=0........(l)

2x-3/2=0............(m)

x/2+5/2=0............(n)

Hence,

$\begin{vmatrix} 2 & 2 & 1/2\\ 2 & 0 & 0\\ 1/2 & 0 & 0 \end{vmatrix}$

= 0;

Hence it is non cental as the determinant is zero.

3. Answer

x²-y²-z² +xy+4yz+x=0

x+y/2+1/2=0

x/2-y+2z=0

2y-z=0

so,

$\begin{vmatrix} 1 & 1/2 & 0 \\ 1/2 & -1 & 2\\ 0 & 2 & -1 \end{vmatrix}$

=1[(-1)(-1)-2*2]-(1/2)[-1*(1/2)-2*0]+0[1/2*2-0(-1)]

=-3-(1/2)[-1/2]

=-3+1/4

=-11/4 != 0

Hence it is a central conicoid.

To find its center,

From ..

2y-z=0

y=z/2......

Put this value in ............

x/2-z/2+2z=0

or x=-3z...........

Put these values of x and y in equation ...

we get

-3z+z/4+1/2=0

or z=2/11

x=-3z=-3*2/11=-6/11

y=z/2=1/2*(2/11)=1/11

hence its center is

(x,y,z)=(-6/11,1/11,2/11)

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