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Cartesian Equation from Parametric x=t²+2t+3, y=t+1 | JEE

JEE Maths question with a full step-by-step solution.

Question
The Cartesian equation of the curve whose parametric equations are x=t2+2t+3x = t^2 + 2t + 3 and y=t+1y = t + 1 is:
Ay=(x−1)2+2(y−1)+3y = (x-1)^2 + 2(y-1) + 3
Bx=(y−1)2+2(y−1)+5x = (y-1)^2 + 2(y-1) + 5
Cx=y2+2x = y^2 + 2correct
Dnone of these
Solution
Step 1: From the second equation t+1=yt + 1 = y. Step 2: Group xx around (t+1)(t+1) and substitute.
x=t2+2t+3=(t+1)2+2=y2+2.x = t^2 + 2t + 3 = (t+1)^2 + 2 = y^2 + 2.
 Correct answer: (3)
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