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The points of intersection of curves whose parametric equations are $x = t^{2} + 1, y = 2 t$ and $x = 2 S, y = \frac{2}{S}$ is

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(1,-3)

(2, 2)

(-2,4)

(1, 2)

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detailed solution

Correct option is B

We have,x=t2+1 and y=2tOn eliminating t, we gety2=4x−4and x=2s and y=2sOn substituting this y2= 4x-4, we get2s3−s2−1=0⇒(s−1)2s2+s+1=0⇒s=1Putting s=1 in x=2s,y=2s, we get x=2, y =2So, point is (2, 2).

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The points of intersection of curves whose parametric equations are x=t2+1,y=2t and x=2S,y=2S is

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The points of intersection of curves whose parametric equations are $x = t^{2} + 1, y = 2 t$ and $x = 2 S, y = \frac{2}{S}$ is