Vertex of the parabola whose parametric equation is $x = t^{2} - t + 1, y = t^{2} + t + 1; t \in R$ , is

(1, 1)

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(2, 2)

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(\frac{1}{2}, \frac{1}{2})

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(3, 3)

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Solution

The correct option is A $(1, 1)$
$x = t^{2} - t + 1, y = t^{2} + t + 1$
$\Rightarrow x + y = 2 (t^{2} + 1)$ and $y - x = 2 t$
$\Rightarrow \frac{x + y}{2} = 1 + {(\frac{y - x}{2})}^{2}$
$\Rightarrow (y - x)^{2} = 2 (x + y) - 4$
$\Rightarrow (y - x)^{2} = 2 (x + y - 2)$
Vertex will be the point where lines $y - x = 0$ and $x + y - 2 = 0$ meet, i.e., the point $(1, 1)$ .

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