### Home > APCALC > Chapter 11 > Lesson 11.2.1 > Problem11-52

11-52.

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

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

1. $x = y^{2} + 1$

1. $x = y + 1$

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

1. $y = x + 1$

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

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