Answer
a. $x=2$ and $D(2)=2$
b. See graph and explanation.
Work Step by Step
a. Step 1. Let $P(x,y)$ be a point on the curve; we have $y=\sqrt {16-x^2}$.
Step 2. The distance $D(x)$ between point P and $(1,\sqrt 3)$ satisfies $D^2=(x-1)^2+(y-\sqrt 3)^2=(x-1)^2+(\sqrt {16-x^2}-\sqrt 3)^2=20-2x-2(48-3x^2)^{1/2}$
Step 3. Let $z=D^2$ and $z'=0$; we have $z'=-2-(48-3x^2)^{-1/2}(-6x)=0$ or $3x=(48-3x^2)^{1/2}$ (need $x\gt0$ here), which gives $x=2$ and $D^2(2)=4, D(2)=2$
Step 4. Check signs of $z'$ across $x=2$ to get $..(-)..(2)..(+)..$; thus the region is concave up with a minimum.
b. See graph; we can identify a minimum on $D(x)$ at $(2,2)$ which agrees with the result from part (a).
