The points of intersection of the curves whose parametric equation are $x = t^{2} + 1$ , $y = 2 t$ and $x = 2 s$ , $y = \frac{2}{s}$ is given by:

(1, - 3)

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(2, 2)

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(- 2, 4)

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(1, 2)

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Solution

The correct option is B $(2, 2)$
For intersection of both the curve we must have,
$t^{2} + 1 = 2 s$ and $2 t = \frac{2}{s} \Rightarrow t s = 1$
$\Rightarrow t^{2} + 1 = \frac{2}{t} \Rightarrow t^{3} + t - 2 = 0$
$\Rightarrow (t - 1) (t^{2} + t + 2) = 0$
$\Rightarrow t = 1$ is the only root
Therefore, the point of intersection is $(2, 2)$
Hence, option 'B' is correct.

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