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If the equations $\mathrm{x}=\mathrm{t}^2+\mathrm{t}+1, \mathrm{y}=\mathrm{t}^2-\mathrm{t}+1$ represents a curve $\mathrm{C}$ with parameter $\mathrm{t}$, then the Cartesian equation of $\mathrm{C}$ is
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Verified Answer
The correct answer is:
$x^2-2 x y+y^2-2 x-2 y+4=0$
Given $\mathrm{x}=\mathrm{t}^2+\mathrm{t}+1, \mathrm{y}=\mathrm{t}^2-\mathrm{t}+1$
Now, $x+y=2\left(t^2+1\right)$ ... (i)
$\mathrm{x}-\mathrm{y}=2 \mathrm{t}$... (ii)
From (i) \& (ii), we get
$\begin{aligned}
& x+y=2\left[1+\left(\frac{x-y}{2}\right)^2\right] \\
& \Rightarrow x+y=\left(\frac{4+x^2+y^2-2 x y}{2}\right) \\
& \Rightarrow x^2+y^2-2 x-2 y-2 x y+4=0
\end{aligned}$
Now, $x+y=2\left(t^2+1\right)$ ... (i)
$\mathrm{x}-\mathrm{y}=2 \mathrm{t}$... (ii)
From (i) \& (ii), we get
$\begin{aligned}
& x+y=2\left[1+\left(\frac{x-y}{2}\right)^2\right] \\
& \Rightarrow x+y=\left(\frac{4+x^2+y^2-2 x y}{2}\right) \\
& \Rightarrow x^2+y^2-2 x-2 y-2 x y+4=0
\end{aligned}$
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