The curve described parametrically by `x = t^2 + t +1`, y = `t^2 - t + 1` represents :

Answer

C

Explanation

Given curves are `x = t^(2) + t + 1` …..(i)
and `y = t^(2) - t + 1` …….(ii)
`x - y = 2t`
Thus, `x = t^(2) + t + 1`
`implies x = ((x - y)/(2))^(2) + ((x - y)/(2)) + 1`
`implies 4x = (x - y)^(2) + 2x - 2y + 4`
`implies (x - y)^(2) = 2 (x + y + 2)`
`implies x^(2) + y^(2) - 2xy - 2x - 2y + 4 = 0`
Now, `Delta = 1.1.4 + 2. (-1)(-1)(-1)-1 xx (-1)^(2) - 1 xx (-1)^(2) - 4 (-1)^(2)`
= 4 - 2 - 1 - 1 - 4 = -4
and `:. Delta =! 0`
and `ab - h^(2) = 1.1. - (-1)^(2) = 1 - 1 = 0`
Hence, it represents a equation of parabola.