$解:设P点的坐标为(4t,\frac {1}{t})$
$所以D点的横坐标为4t,C点的纵坐标为\frac {1}{t}$
$因为点C,D在反比例函数y=\frac {1}{x}上$
$所以D(4t,\frac {1}{4t}),C(t,\frac {1}{t})$
$所以PC=3t,PD=\frac {1}{t}-\frac {1}{4t}=\frac {3}{4t}$
$S_{△PCD}=\frac {1}{2}×PC×PD=\frac {1}{2}×3t×\frac {3}{4t}=\frac {9}{8}$