............................…···8分\r将点 M 的坐标代入，可得 h= 4h + (3 - 3a)． 解得 h=a-1.\r而 BH —OH —0B =a — l，从而总有 h= BH. ·············--·········…....….................…· 10分\r1\r＠由O知， 点M的坐标为(2a-2,a-1)， 点N的坐标为（a，了1)．\rl l l l l 3a- 3\rS=S凶ONH-S 60Nc = -:-NH x OH --:-OG x h =-:-x-:-axa--:-x- —x(a- 1)\r2 2 2 2 2 2\r1 ? 3 3 1 (3 \2 3\r= - ½a2 +%a - ¾= - ½(a - %J +¾. •···································································· 12分\r2 2 4 2 2 8\r3 3\r当 a=— 时 ， S有最大值 ， 最大值为 — .\r2 8\r3 3\rS取最大值时点P的坐标为（一—, ]···············. ··························································14分\r2 2
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