Answer
$L(y)=\sqrt{y^4-y^2+1}$
Work Step by Step
Let's say, $\ (x,y)\ $ is some random point on the graph of the given function $\ y=\sqrt{x-3}.\ $ Since this random point is on the graph of the function $\ y=\sqrt{x-3}\ $, we can write the coordinates of it as, $\ (x,\sqrt{x-3}).$
Now use the distance formula to find the distance between point $\ (x,\sqrt{x-3})\ \ \mathrm{and}\ \ (4,0).$
$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\Rightarrow\ L=\sqrt{(4-x)^2+(0-\sqrt{x-3})^2}$
$\Rightarrow\ L=\sqrt{16-8x+x^2+x-3}$
$\Rightarrow\ L=\sqrt{x^2-7x+13}$
Now write the given function in terms of $\ y\ $.
$\ y=\sqrt{x-3}$
$\Rightarrow\ y^2=x-3$
$\Rightarrow\ x=y^2+3$
Put this obtained function into $\ L\ $ to get:
$L=\sqrt{(y^2+3)^2-7(y^2+3)+13}$
$\Rightarrow\ L=\sqrt{y^4+6y^2+9-7y^2-21+13}$
$\Rightarrow\ L=\sqrt{y^4-y^2+1}$
$L(y)=\sqrt{y^4-y^2+1}$
