14. Let the points on the curve $2 x^{2}+y^{2}=4 x+6$ that are farthest from and closest to the origin be $M$ and $N$, respectively. Then $|M N|=$ $\qquad$ .
14. $\sqrt{15}$.

Given $y^{2}=-2 x^{2}+4 x+6 \geqslant 0$, we get $-1 \leqslant x \leqslant 3$.
Take any point $P(x, y)$ on the curve $2 x^{2}+y^{2}=4 x+6$, the distance from $P$ to the origin is
$$
\begin{array}{l}
d=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}-2 x^{2}+4 x+6} \\
=\sqrt{-(x-2)^{2}+10} .
\end{array}
$$

Since $-1 \leqslant x \leqslant 3$, we have $1 \leqslant d \leqslant \sqrt{10}$.
When $x=2$, $d$ reaches its maximum value $\sqrt{10}$, then $M(2, \sqrt{6})$ or $M(2,-\sqrt{6})$;
When $x=-1$, $d$ reaches its minimum value 1, then $N(-1,0)$.
Thus, $|M N|=\sqrt{(2+1)^{2}+( \pm \sqrt{6})^{2}}=\sqrt{15}$.