• # question_answer The points of intersection of the curve whose parametric equations are $x={{t}^{2}}+1$ and $y=2t$and$x=2s,\,\,y=\frac{2}{s}$ is given by A)  $(1,\,\,-3)$                  B)  $(2,\,\,2)$ C)  $(-2,\,\,4)$                  D)  $(1,\,\,2)$

$\because$    $x={{t}^{2}}+1$ and $y=2t$ $\Rightarrow$            $x={{\left( \frac{y}{2} \right)}^{2}}+1$ $\Rightarrow$            ${{y}^{2}}=4(x-1)$ Also,    $x=2s$ and $y=2/s$ $\therefore$    $xy=4$ From Eqs. (i) and (ii) we get $x=2$ $\Rightarrow$            $y=\frac{4}{2}=2$ Hence, point is (2, 2).