Find the Cartesian equation of the curve whose parametric equations are $x = t^{2} + t + 1$ , $y = t^{2} - t + 1$ .

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Solution

$\frac{d y}{d x} = 2 t + 1 \frac{d y}{d t} = 2 t - 1 \frac{d y}{d t} \times \frac{d t}{d x} = \frac{2 t - 1}{2 t + 1} \frac{d y}{d x} = \frac{2 t - 1}{2 t + 1} \int d y = \int \frac{2 t - 1}{2 t + 1} d t \times \frac{d x}{d t} y = \int \frac{2 t - 1}{2 t + 1} \times (2 t + 1) d t y = \int (2 t - 1) d t$
$y = t^{2} - t + 1$
$y = (t^{2} + 1) - t$
$y = x - t - t$
$y = x - 2 t$

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