14. $x = - \frac { 3 } { 2 }$
Solution: The focus has coordinates $F \left( \frac { p } { 2 } , 0 \right)$, and point $P$ has coordinates $P \left( \frac { p } { 2 } , p \right)$. Thus $| P F | = p , | O F | = \frac { p } { 2 }$. By the focal chord property $| P F | ^ { 2 } = | O F | \cdot | F Q |$, we get $p ^ { 2 } = 3 p$, so $p = 3$. The directrix equation is $x = - \frac { 3 } { 2 }$.
15. 1
Solution: When $x \geq \frac { 1 } { 2 }$, $f ( x ) = 2 x - 2 \ln ( x ) - 1$, and $f ^ { \prime } ( x ) = \frac { 2 ( x - 1 ) } { x } > 0$, so $f ( x ) \geq f ( 1 ) = 1$. When $x < \frac { 1 } { 2 }$, $f ( x ) = 1 - 2 \ln ( x ) - 2 x$, and $f ^ { \prime } ( x ) = \frac { - 2 ( x + 1 ) } { x } < 0$, so $f ( x ) > f \left( \frac { 1 } { 2 } \right) = 2 \ln 2 > 1$. Therefore, the minimum value of function $f ( x )$ is 1.