Question

If the equations \( x = t^2 + t + 1, \quad y = t^2 - t + 1 \) represent a curve \( C \) with parameter \( t \), then the Cartesian equation of \( C \) is:

• A. \( x^2 - 2xy + y^2 - 2x - 2y + 4 = 0 \)
• B. \( x^2 + 2xy + y^2 - 2x - 2y + 4 = 0 \)
• C. \( x^2 - 2xy + y^2 + 2x + 2y + 4 = 0 \)
• D. \( x^2 - 2xy - y^2 + 2x + 2y + 4 = 0 \)

Hint:

Eliminate parameter \( t \) using \( x \pm y \) to get expressions for \( t \) and \( t^2 \), then substitute and simplify.

Answer 1

The Correct Option is A

Given: \[ x = t^2 + t + 1,\quad y = t^2 - t + 1 \] Add: \[ x + y = 2t^2 + 2 \Rightarrow t^2 = \dfrac{x + y - 2}{2} \] Subtract: \[ x - y = 2t \Rightarrow t = \dfrac{x - y}{2} \] Now express in terms of \( x \) and \( y \): \[ t = \dfrac{x - y}{2},\quad t^2 = \dfrac{x + y - 2}{2} \] Now substitute into: \[ x = t^2 + t + 1,\quad y = t^2 - t + 1 \] Using above values and simplifying gives the final Cartesian equation: \[ x^2 - 2xy + y^2 - 2x - 2y + 4 = 0 \] % Final Answer \[ \boxed{x^2 - 2xy + y^2 - 2x - 2y + 4 = 0} \]