Question

If the systems of equations
$ax+hy+g=0$ (1)
$hx+by+f=0$ (2)
and $a{x}^2+2hxy+b{y}^2+2gx+2fy+c+t=0$ (3)
has a unique solution, and
$\frac{abc+2fgh-a{f}^2-b{g}^2-c{h}^2}{h^2-ab}=8,$ find t

Answer 1

Given system of equations
$ax+hy+g=0$ (1)
$hx+by+f=0$ (2)
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c+t=0$ (3)
Since given system has a unique solution.
$\Rightarrow ab-h^2\ne 0$.
We can rewrite (3) as
$x(ax+hy+g)+y(hx+by+f)+gx+fy+c+t=0$ (4)
Since any solution of (1) and (2) must be a solution of(3), the given system of equations can be rewritten as
$ax+hy+g=0,hx+by+f=0,gx+fy+c+t=0$
Since the above system of equations is consistent, we must have
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c+t\end{array}\right|=0$
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|+\left|\begin{array}{ccc}a& h& 0\\ h& b& 0\\ g& f& t\end{array}\right|=0$
$\Rightarrow [abc+2fgh-a{f}^2-b{g}^2-c{h}^2]+t(ab-h^2)=0$
$\Rightarrow t=\frac{abc+2fgh-a{f}^2-b{g}^2-c{h}^2}{h^2-ab}$
$\Rightarrow t=8$