The conic section represented by the equations $x = t^{2} + t + 1$ and $y = t^{2} - t + 1$ is ________ .

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Solution

Given equations are $x = t^{2} + t + 1$ and $y = t^{2} - t + 1$ .
$x + y = 2 t^{2} + 2$ and $x - y = 2 t$
$x + y = 2 t^{2} + 2 \Rightarrow 2 (x + y) = {(2 t)}^{2} + 4 \Rightarrow 2 (x + y) = {(x - y)}^{2} + 4 \Rightarrow x^{2} + y^{2} - 2 xy - 2 x - 2 y + 4 = 0$
Here, a = 1, b = 1, h = -1, g = -1, f = -1, c = 4.
Now, $△ = abc + 2 fgh - {af}^{2} - {bg}^{2} - {ch}^{2} \Rightarrow △ = - 4 \neq 0$ And $h^{2} = ab$
So, the given equation represents a parabola.

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