Correct Answer - B
We have,
`x=t^(2)+t+1, y=t^(2)-t+1`
`rArr" "x+y=2(t^(2)+1)" and, "x-y=2t`
`rArr" "x+y=2{((x-y)/2)2+1}`
`rArr" "2(x+y)=(x-y)^(2)+4`
`rArr" "x^(2)+y^(2)-2xy-2y-2y+4=0`
Comparing this equation with the equation
`ax^(2)+2hxy+by^(2)+2gx+2fy+c=0` we get
`a=1, b=1, c=4, h=-1, g=-1, " and " f=-1`
`:." "abc+2fgh-af^(2)-bg^(2)-ch^(2)=4-2-1-1-4ne0" and, "h^(2)-ab=1-1=0`
Thus, we have
`Deltane0" and "h^(2)=ab`
So, the given equations represetn a parabola.