1. The line $l$ passes through the focus of the parabola $y^{2}=a(x+1)(a>0)$ and is perpendicular to the $x$-axis. If the segment cut off by $l$ on the parabola is 4, then $a=$ $\qquad$ .
$=.1 .4$.
Since the parabolas $y^{2}=a(x+1)$ and $y^{2}=a x$ have the same length of the focal chord with respect to their directrix, the general equation $y^{2}=a(x+1)$ can be replaced by the standard equation $y^{2}=a x$ for solving, and the value of $a$ remains unchanged. Using the formula for the length of the latus rectum, we get $a=4$.