The Locus of a point whose co_ordinate are by $x = t + t^{2}, y = 2 t + 1$ where $t$ is variable is

y^{2} = 4 x + 1

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y^{2} = 4 (x + 1)

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y^{2} = 4 x

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Solution

The correct option is A $y^{2} = 4 x + 1$
We have,
$\begin{matrix} x = t + t^{2} & y = 2 t + 1 = t (t + 1) & y - 1 = 2 t x = \frac{y - 1}{2} (\frac{y - 1}{2} + 1) 4 x = (y - 1) (y + 1) 4 x = y^{2} y^{2} = 4 x + 1 \end{matrix}$

Hence, this is the answer.

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