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11-51.

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:

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