Answer to Question #86622 in Analytic Geometry for RAKESH DEY

Question #86622
Examine which of the following conicoids are central and which are non-central. Also determine which of the central conicoids have centre at the origin.
1) x^2+y^2+z^2+x+y+z=1
2) 2x^2+4xy+xz-x-3y+5z+3=0
3) x^2-y^2-z^2+xy+4yz+x=0
1
Expert's answer
2019-03-25T14:54:44-0400

A conicoid given by equation ax2+by2+cz2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0 has a point (x0,y0,z0) as a center if


"\\begin{cases}\nax_0+hy_0+gz_0+u=0 \\\\\nhx_0+by_0+fz_0+v=0 \\\\\ngx_0+fy_0+cz_0+w=0\n\\end{cases}"

For 1) a=b=c=1, u=v=w=1/2, d=-1, f=g=h=0:


"\\begin{cases}\nx_0+1\/2=0 \\\\\ny_0+1\/2=0 \\\\\nz_0+1\/2=0\n\\end{cases}"

x2+y2+z2+x+y+z=1 - the central conicoid have a center (-1/2,-1/2,-1/2)


For 2) a=2, h=2, g=1/2, u=-1/2, v=-3/2, w=5/2, d=3, b=c=f=0:


"\\begin{cases}\n2x_0+2y_0+z_0\/2-1\/2=0 \\\\\n2x_0-3\/2=0 \\\\\nx_0\/2+5\/2=0\n\\end{cases}"

2x2+4xy+xz-x-3y+5z+3=0 - the conicoid not central


For 3) a=1, b=c=-1, h=1/2, f=2, u=1/2, d=g=v=w=0:


"\\begin{cases}\nx_0+y_0\/2+1\/2=0 \\\\\nx_0\/2-y_0+2z_0=0 \\\\\n2y_0-z_0=0\n\\end{cases}"

"\\begin{cases}\nx_0=-1\/2-z_0\/4 \\\\\n-1\/4-z_0\/8-z_0\/2+2z_0=0 \\\\\ny_0=z_0\/2\n\\end{cases}"

"\\begin{cases}\nx_0=-6\/11 \\\\\nz_0=2\/11 \\\\\ny_0=1\/11\n\\end{cases}"

x2-y2-z2+xy+4yz+x=0 - the central conicoid have a center (-6/11,1/11,2/11)



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