The Cartesian equation of the curve whose parametric equations are $x = t^{2} + 2 t + 3$ and y = t + 1 is

y = (x - 1)^{2} + 2 (y - 1) + 3

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x = (y - 1)^{2} + 2 (y - 1) + 5

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x = y^{2} + 2

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none of these

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Solution

The correct option is C $x = y^{2} + 2$
$x = t^{2} + 2 t + 3 = (t + 1)^{2} + 2 = y^{2} + 2$

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Similar questions

x = t^{2} + 2 t + 3

x = t^{2} + t + 1

y = t^{2} - t + 1

The points of intersection of the curves whose parametric equations are $x = t^{2} + 1, y = 2 t a n d x = 2 s, y = \frac{2}{s}$ Is given by

The point of intersection of the curves whose parametric equations are $x = t^{2} + 1, y = 2 t$ and $x = 2 s, y = 2 / s$ is given by

x = t^{2} + 1

y = 2 t

x = 2 s

y = \frac{2}{s}

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