14. Let the curve $C:\left|x^{2}-16 y\right|=256-16|y|$ enclose a closed region $D$.
(1) Find the area of region $D$; (2) Let a line passing through the point $M(0,-16)$ intersect the curve $C$ at two points $P, Q$, find the maximum value of $|P Q|$.
Solution (1) From the given, we have $256-16|y| \geq 0$, thus $-16 \leq y \leq 16$.
If $\left|x^{2}-16 y\right|=x^{2}-16 y$, then when $0 \leq y \leq 16$, $\left|x^{2}-16 y\right|=x^{2}-16 y=256-16 y, x^{2}=256$, at this time $x= \pm 16(0 \leq y \leq 16)$, the image is two line segments;
When $-16 \leq y \leq 0$, $\left|x^{2}-16 y\right|=x^{2}-16 y=256+16 y, y=\frac{x^{2}}{32}-8(y \geq-8)$, corresponding to a segment of a quadratic function image; if $\left|x^{2}-16 y\right|=16 y-x^{2}$, then when $0 \leq y \leq 16$, similar to the previous derivation, we get $y=\frac{x^{2}}{32}+8$, corresponding to a segment of a quadratic function image: $y=\frac{x^{2}}{32}+8(y \geq 8)$;
When $-16 \leq y<0$, $\left|x^{2}-16 y\right|=16 y-x^{2}=256+16 y$, we get $x^{2}=-256$, no solution.
In summary, the set of region $D$ is: $D=\left\{(x, y) \mid-16 \leq x \leq 16, \frac{x^{2}}{32}-8 \leq y \leq \frac{x^{2}}{32}+8\right\}$, from the properties of the function image on region $D$, the area of region $D$ is $S=32 \times 16=512$.
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(2) Let the line passing through point $M(0,-16)$ be $l$. To find the maximum value of $|P Q|$, by the symmetry of region $D$, we only need to consider the part of the image of $D$ on the right side of the $y$-axis that intersects with line $l$.
Let the equation of the line $l$ passing through point $M(0,-16)$ be $y=k x-16$, it is easy to know that at this time $l$ intersects with $D$ when $1 \leq k<\infty$.
(1) When $2 \leq k<\infty$, $l$ intersects with $D$ at the quadratic functions $y=\frac{x^{2}}{32}-8$ and $y=\frac{x^{2}}{32}+8$, the two intersection points are $P\left(16\left(k-\sqrt{k^{2}-1}\right), 16\left(k^{2}-k \sqrt{k^{2}-1}-1\right)\right), Q\left(16\left(k-\sqrt{k^{2}-3}\right), 16\left(k^{2}-k \sqrt{k^{2}-3}-1\right)\right)$, thus $|P Q|=$ $16\left(\sqrt{k^{2}-1}-\sqrt{k^{2}-3}\right) \sqrt{1+k^{2}}$, which is a decreasing function of $k$;
(2) When $1 \leq k \leq 2$, the line $l$ intersects with $D$ at the quadratic function $y=\frac{x^{2}}{32}-8$ and the line $y=16$, from the properties of the image, it is easy to see that as $k$ changes from 2 to 1, the value of $|P Q|$ gradually decreases.

In summary, when $l$ passes through the intersection point $Q(16,16)$ of the line $x=16$ and the curve of the quadratic function $y=\frac{x^{2}}{32}+8$, the value of $|P Q|$ is maximized, at this time the equation of the line $l$ is: $y=2 x-16, P(16(2-\sqrt{3}), 16(3-2 \sqrt{3})),|P Q|$ is
$$
\sqrt{[16(3-2 \sqrt{3})-16]^{2}+[16(2-\sqrt{3})-16]^{2}}=16 \sqrt{20-10 \sqrt{3}}
$$

When $|P Q|$ lies on the $y$-axis, $|P Q|=24<16 \sqrt{20-10 \sqrt{3}}$, therefore, the maximum value of $|P Q|$ is $16 \sqrt{20-10 \sqrt{3}}$.