The points of intersection of the curves whose parametric equation are x=t2+1, y=2t and x=2s, y=2s is given by:
A
(1,−3)
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B
(2,2)
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C
(−2,4)
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D
(1,2)
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Solution
The correct option is B(2,2) For intersection of both the curve we must have, t2+1=2s and 2t=2s⇒ts=1 ⇒t2+1=2t⇒t3+t−2=0 ⇒(t−1)(t2+t+2)=0 ⇒t=1 is the only root Therefore, the point of intersection is (2,2) Hence, option 'B' is correct.