For the hyperbola 9x2−16y2−18x+32y−151=0

The given hyperbola can be written as(x−1)216−(y−1)29=1or X216−Y29=1 (where X=x−1,Y=y−1 ) e=1+b2a2=1+916=54Directrices areX=±a/e. Therefore, x−1=±165or x=215 and x=−115Length of latus rectum =2b2a=92The foci are given byX=±ae,Y=0 or  (6,1),(−4,1)