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11-52.

Let $x =\sqrt { 4 t ^ { 2 } + 1 }$ , $y = 2t$ , and $t > 0$ . An equivalent rectangular equation is:

$x =\sqrt { y ^ { 2 } + 1 }$

$x = y^{2} + 1$

$x = y + 1$

$y =\sqrt { x ^ { 2 } + 1 }$

$y = x + 1$

Step 1:

$t=\frac{y}{2}$

Step 2:

$x=\sqrt{4\Big(\frac{y}{2}\Big)^2+1}$

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