Question

The number of integral values of $k$ for which the equation $(k+1)x^2+2(k-1)xy+y^2-x+2y+3=0$ represents an ellipse, is

Answer 1

The general equation of second degree is :
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
Given : $(k+1)x^2+2(k-1)xy+y^2-x+2y+3=0$
Comparing both,
$a=(k+1),b=1,c=3,f=1,g=-\frac{1}{2},h=k-1$

For conics to be an ellipse,
$h^2<ab$
$\Rightarrow (k-1{)}^2<k+1$
$\Rightarrow k^2-3k<0$
$\Rightarrow k(k-3)<0$
$\Rightarrow k\in (0,3)$
Possible integral values of $k$ are $1$ and $2$

Also, $\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$\Rightarrow \Delta =3(k+1)-(k-1)-(k+1)-\frac{1}{4}-3(k-1{)}^2$
$\Delta \ne 0$ for $k=1$ or $k=2$

$\therefore$ Possible integral values of $k$ are $1,2$