I have the following second degree equation in $x,y$ & $z$: $x^2-2y^2+3z^2+5yz-6xz-4xy+8x-19y-2z-20=0$ I calculated the discriminating cubic as: $k^3-4k^2-97k-190=0$
The roots of this equation are distinct. The question says that it represents a cone having vertex at$(1,-2,3)$.
As a cone is a surface of revolution, two roots of the discriminating cubic should be zero. But that's not the case here. So this equation can't represent a cone.
Am I correct in asserting this? Can the equation represent a cone even without having two equal roots? Please suggest.
if you take $a=x-1, b = y+2, c=z-3$ you arrive at $$ a^2 - 2 b^2 + 3 c^2 + 5 bc -6 ca -4 ab = 0. $$ It is this $0$ on the right hand side that says the object is a cone (over an ellipse, not a circle). Think about it: if you had $x^2 + y^2 - z^2 = 0$ you would have a circular cone. if you had $x^2 + 2y^2 - z^2 = 0$ you would have an elliptical cone, as each choice of $z$ leads to a constant value for $x^2 + 2 y^2$
The fact that we get a quadratic form equal to zero comes from the final diagonal element $0$ in $D$ in these expressions:
$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ \frac{ 11 }{ 6 } & - \frac{ 7 }{ 12 } & 1 & 0 \\ 1 & - 2 & 3 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & - 4 & - 6 & 8 \\ - 4 & - 4 & 5 & - 19 \\ - 6 & 5 & 6 & - 2 \\ 8 & - 19 & - 2 & - 40 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 2 & \frac{ 11 }{ 6 } & 1 \\ 0 & 1 & - \frac{ 7 }{ 12 } & - 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - 12 & 0 & 0 \\ 0 & 0 & - \frac{ 95 }{ 12 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 2 & 1 & 0 & 0 \\ - 3 & \frac{ 7 }{ 12 } & 1 & 0 \\ 4 & \frac{ 1 }{ 4 } & - 3 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - 12 & 0 & 0 \\ 0 & 0 & - \frac{ 95 }{ 12 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - 2 & - 3 & 4 \\ 0 & 1 & \frac{ 7 }{ 12 } & \frac{ 1 }{ 4 } \\ 0 & 0 & 1 & - 3 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & - 4 & - 6 & 8 \\ - 4 & - 4 & 5 & - 19 \\ - 6 & 5 & 6 & - 2 \\ 8 & - 19 & - 2 & - 40 \\ \end{array} \right) $$
We can parametrize all integer solutions in four ways as $$ \begin{array}{} a=3u^2 +5uv-2v^2 \\ b=-2u^2 +4uv+v^2 \\ c=u^2 +2uv-5v^2 \; . \end{array} $$
OR $$ \begin{array}{} a=7uv-4v^2 \\ b=-u^2 +2uv+5v^2 \\ c=2u^2 +4uv-v^2 \; . \end{array} $$ OR $$ \begin{array}{} a=7uv-8v^2 \\ b=3u^2 -6uv+v^2 \\ c=u^2 +4uv-2v^2 \; . \end{array} $$
OR$$ \begin{array}{} a=2u^2 +uv-6v^2 \\ b=u^2 +6uv+3v^2 \\ c=3u^2 -2v^2 \; . \end{array} $$
in the following, I named the four (u,v) recipes with letters A,B,C,D. If the triple produced had $a < 0,$ I printed $-a,-b,-c$