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Q. The curve described parametrically by $x = t^2 + 2t - 1, y = 3t + 5$ represents

VITEEEVITEEE 2007

Solution:

Given $x = t^2 + 2t-1 \; \& \; y = 3t + 5$
$\Rightarrow \; x = t^2 + 2t + 1-2 \; \& \; y = 3t + 3+2$
$\Rightarrow \; x = (t+1)^2 -2 \; \& \; y = 3(t + 1) + 2 (2)$
$\Rightarrow (t+1) = \sqrt{ x + 2}$ .......... (1)
$\therefore $ Equation (2) becomes [using equation (1)]
$y = 3 \sqrt{ x + 2} + 2$
$\Rightarrow \; y - 2 = 3 \sqrt{x +2}$
squaring both sides, we get
$(y-2)^2 = 9 (x + 2)$
$\Rightarrow \; Y^2 =9X$ where $Y = y - 2 \; \& \; X = x + 2$
This equation represents a parabola