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 $(1, 1)$
$x = t^{2} - t + 1, y = t^{2} + t + 1$
$x + y = 2 (t^{2} + 1) and y - x = 2 t$
$\Rightarrow \frac{x + y}{2} = 1 + {(\frac{y - x}{2})}^{2}$
$\Rightarrow (y - x)^{2} = 2 (x + y) - 4$
$\Rightarrow (y - x)^{2} = 2 (x + y - 2)$
Vertex will be the point where the lines
$y - x = 0 and x + y - 2 = 0$ meets,
$x + x - 2 = 0 \Rightarrow x = 1 \Rightarrow y = 1$
Hence the vertex is $(1, 1)$

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