A determinant and a locus - 2

The determinant calculates to $3xy + x + 10y - y^2 - 5x^2 - 6 = 0$ , which after rearranging is $5x^2 - 3xy + y^2 - x - 10y + 6 = 0$ .

Using the equations found here , $A = 5$ , $B = -3$ , $C = 1$ , $a = 5$ , $b = 1$ , $c = 6$ , $f = -5$ , $g = -\frac{1}{2}$ , and $h = -\frac{3}{2}$ .

Therefore, $abc + 2fgh - af^2 - bg^2 - ch^2 = -\frac{465}{4} \neq 0$ , so the conic is non-degenerate, and $B^2 - 4AC = -11 < 0$ and $A \neq C$ , so the conic is an ellipse .

Expanding, the equation is $5x^2 - 3xy + y^2 - x - 10y - 6 \; = \; 0$ We can find $u,v,w$ such that the equation becomes $5(x-u)^2 - 3(x-u)(y-v) + (y-v)^2 \; = \; w$ and we can find an orthgonal matrix $P$ and real numbers $p,q$ such that the equation becomes $pX^2 + qY^2 \; = \; w \hspace{2cm} \binom{X}{Y} \; = \; P\binom{x-u}{y-v}$ and $p,q$ are the eigenvalues of the matrix $\left(\begin{array}{cc} 5 & -\frac32 \\ -\frac32 & 1 \end{array}\right)$ This matrix has characteristic polynomial $t^2 - 6t + \tfrac{11}{4}$ , which has two positive real roots. Thus the locus is either an ellipse (if $w > 0$ ) or the empty set (if $w < 0$ ) or a single point (if $w=0$ ). Note that $(6,18)$ is a solution of this equation (this makes the first and third columns parallel) and also $(\tfrac15,10)$ is a solution (first and second columns parallel). Thus we have an ellipse.