$∵四边形OABC是平行四边形$
$∴BC//AO$
$S_{平行四边形OABC}=2S_{△OBC}$
$∴点B的纵坐标为\frac {6}{a}$
$∵OB的解析式为y=\frac23x$
$∴B(\frac {9}{a},\frac {6}{a})$
$∴BC=\frac {9}{a}-a$
$∴S_{△OBC}=\frac12×\frac {6}{a}×(\frac {9}{a}-a)$
$∴2×\frac12×\frac {6}{a}×(\frac {9}{a}-a)=\frac {15}{2}$
$解得a=2或-2(舍)$
$∴B(\frac {9}{2},3)$