**General Equation of the Second Degree**

The surface represented by a general equation of the second degree in x, y, z is called a second-order surface.

The general equation of the second degree in x, y, z is usually written in the form:

\[a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2fyz+2gzx+2ux+2vy+2wz+d=0………\left( 1 \right)\]

Hence (1) represents a second order surface. Such a surface is sometimes called a quadratic surface or simply a quadratic or a **conicoid**.

Then S=-u\left( Au+Hv+Gw \right)-v\left( Hu+Bv+Fw \right)-w\left( Gu+Fv+Cw \right)+d\Delta

Where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h respectively in △.

A complete classification of quadrics according to different possible values of △ and S, the corresponding canonical forms and their corresponding names are given bellow in tabular form.

△ | S | Canonical form | Name |

≠ 0 | < 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}+\frac{{{z}^{2}}}{{{\gamma }^{2}}}=1\] | Ellipsoid |

≠ 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}+\frac{{{z}^{2}}}{{{\gamma }^{2}}}=0\] | Point-ellipsoid |

≠ 0 | > 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}+\frac{{{z}^{2}}}{{{\gamma }^{2}}}=1\] | No Geometric Locus |

≠ 0 | > 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}-\frac{{{z}^{2}}}{{{\gamma }^{2}}}=1\] | Hyperboloid of one sheet |

≠ 0 | < 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}-\frac{{{y}^{2}}}{{{\beta }^{2}}}-\frac{{{z}^{2}}}{{{\gamma }^{2}}}=1\] | Hyperboloid of two sheets |

\[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}-\frac{{{z}^{2}}}{{{\gamma }^{2}}}=0\]≠ 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}-\frac{{{z}^{2}}}{{{\gamma }^{2}}}=0\] | Quadric Cone |

= 0 | > 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}=2cz\] | Elliptic Paraboloid |

= 0 | < 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}-\frac{{{y}^{2}}}{{{\beta }^{2}}}=2cz\] | Hyperbolic Paraboloid |

= 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}=1\] | Elliptic Cylinder |

= 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}-\frac{{{y}^{2}}}{{{\beta }^{2}}}=1\] | Hyperbolic Cylinder |

= 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}=-1\] | Non Geometric locus (imaginary cylinder) |

= 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}+\frac{{{y}^{2}}}{{{\beta }^{2}}}=0\] | Non Geometric Locus (pair of imaginary planes with real intersection) |

= 0 | = 0 | \[\frac{{{x}^{2}}}{{{\alpha }^{2}}}-\frac{{{y}^{2}}}{{{\beta }^{2}}}=0\] | Pair of Intersecting planes |

= 0 | = 0 | \[{{x}^{2}}=2py,\,\,p\ne 0\] | Parabolic Cylinder |

= 0 | = 0 | \[{{x}^{2}}-{{\alpha }^{2}}=0,\,\,\alpha \ne 0\] | Pair of parallel planes |

= 0 | = 0 | \[{{x}^{2}}+{{\alpha }^{2}}=0,\,\,\alpha \ne 0\] | No Geometric Locus (Pair of imaginary parallel planes) |

= 0 | = 0 | \[{{x}^{2}}=0\] | Pair of Coincident planes |

Example 01 |

**For each of the following quadrics determine whether it has a single center, no center, a line of centers or a plane of centers.**

**\[\left( i \right)\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2xy-2yz-2zx-2x-4y-2z-3=0\]**

**\[\left( ii \right)\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2xy-2yz+2zx-2x-4y-2z+3=0\]**

**Solution:**

(i) Comparing the given equation with the general equation of the second degree in x, y, z

\[a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2fyz+2gzx+2ux+2vy+2wz+d=0\]

We have, a=1,b=1,c=1,h=-1,g=-1,f=-1,f=-1,u=-1,v=-2,w=-1,d=-3

We now consider the system of equations

ax+hy+gz+u=0,\,\,hx+by+fz+v=0,\,\,gx+fy+cz+w=0 which takes the form

\[x-y-z-1=0,\,\,-x+y-z-2=0,\,\,-x-y+z-1=0\]

Hence the above system of equations has a unique solution namely \left( -\frac{3}{2},-1,-\frac{3}{2} \right). So the given quadric has a single center namely \left( -\frac{3}{2},-1,-\frac{3}{2} \right).

(ii) Comparing the given equation with the general equation of the second degree in x, y, z

\[a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2fyz+2gzx+2ux+2vy+2wz+d=0\]

We have, a=1,b=1,c=1,h=-1,g=1,f=-1,f=-1,u=-1,v=-2,w=-1,d=3

We now consider the system of equations

ax+hy+gz+u=0,\,\,hx+by+fz+v=0,\,\,gx+fy+cz+w=0 which takes the form x-y+z-1=0,\,\,-x+y-z-2=0,\,\,x-y+z-1=0

Since the first and the third equations of the system are the same, we have to consider the system given by

\[x-y+z-1=0,\,\,-x+y-z-2=0\]

\[i.e.,\,\,the\,\,system\,\,x-y+z-1=0,\,\,x-y+z-2=0\]

This system has no solution as the planes represented by them are parallel.

Hence the given quadric has no center.

Example 02 |

**Find the nature of the quadric represented by \[{{y}^{2}}+4zx+2x-4y+6z+5=0\]**

**Solution:**

Comparing the given equation with the general equation of the second degree in x, y, z

\[a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2fyz+2gzx+2ux+2vy+2wz+d=0\]

We have, a=0,b=2,c=0,h=0,g=2,f=0,f=-1,u=1,v=-2,w=3,d=5

Hence the given quadric represents a cone.

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