Answer
$L= \sqrt {y^4 - y^2 +1}$
Work Step by Step
$y= \sqrt {x-3}$
$y^2= x-3$
$x = y^2 + 3$
Say a is the distance between (x,0) and (4,0), then a = x-4
Say b is the distance between (4,y) and (4,0) then b = y-0
Using Pythagorean's theorem, one can say that $L^2 = a^2 + b^2$. Thus $L = \sqrt {a^2 + b^2}$.
So, $L = \sqrt {(x-4)^2 + (y-0)^2}$.
Substituting $x = y^2 + 3$ gives
$L = \sqrt {(y^2 + 3-4)^2 + (y-0)^2}$
$L = \sqrt {(y^2 -1)^2 + y^2}$
$L = \sqrt {y^4 -2y^2 +1 + y^2}$
$L = \sqrt {y^4 -y^2 +1}$
