Answer
a. $x=1$ and $D=\frac{\sqrt 5}{2}$
b. See graph and explanation.
Work Step by Step
a. Step 1. Let $P(x,y)$ be a point on the curve; we have $y=\sqrt x$.
Step 2. The distance $D(x)$ between point P and $(3/2,0)$ satisfies $D^2=(x-\frac{3}{2})^2+y^2=(x-\frac{3}{2})^2+x$ and $D=\sqrt {(x-\frac{3}{2})^2+x}$. Let $z=D^2$ and $z'=0$; we have $z'=2(x-\frac{3}{2})+1=0$, which gives $x=1$ and $D=\frac{\sqrt 5}{2}$
Step 3. Check $z''=2\gt0$; thus the region is concave up with a minimum.
b. See graph. We can identify a minimum on $D(x)$ at $(1,\frac{\sqrt 5}{2})$, which agrees with the result from part (a).
