Question

If the equation $k{y}^2+y=x^2-16x+64$ represents a parabola then the value of $|4\Delta |=$

Answer 1

The given equation is $k{y}^2+y=x^2-16x+64$
compairing it with standard conic equation,
we get $a=1,b=-k,h=0,g=-8,f=-1/2,c=64$
For equation to be parabola
$h^2=ab\Rightarrow k=0$
Now
$\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0+0-1/4=-\frac{1}{4}\Rightarrow |4\Delta |=1$