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

(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)

Eliminating t from x = $t^{2}$ + 1, y = 2t, we obtain $y^{2} = 4 x - 4$

Similarly eliminating s from x = 2s, y = $\frac{2}{s}$ , We get xy = 4.
Solving these two equations, we get the point the point of intersection.

Hence point of intersection is (2, 2).

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Similar questions

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}

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

x = t^{2} + t + 1

y = t^{2} - t + 1

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