Question 1173368
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<pre>

From  x = {{{1/sqrt(t+1)}}},  you get, squaring,


    x^2 = {{{1/(t+1)}}},

    x^2*(t+1) = 1

    x^2*t + x^2 = 1

    x^2*t = 1 - x^2

    t = {{{(1-x^2)/x^2}}}.


Now substitute it into the expression for y = {{{t/(t+1)}}}.  You will get


    y = {{{(1-x^2)/x^2)}}} : {{{((1-x^2)/x^2 + 1)}}} = {{{(1-x^2)/x^2)}}} : {{{1/x^2}}} = 1- x^2.


So, the curve is the parabola


    y = 1- x^2.      <U>ANSWER</U>
</pre>

Solved.