摘要：088]解:(1)法一:由图象可知:抛物线经过原点. 设抛物线解析式为． 把.代入上式得:···················································································· 1分 解得······················································································· 3分 ∴所求抛物线解析式为········································································ 4分 法二:∵.. ∴抛物线的对称轴是直线． 设抛物线解析式为()······························································ 1分 把.代入得 解得··········································································· 3分 ∴所求抛物线解析式为．····························································· 4分 (2)分三种情况: ①当.重叠部分的面积是.过点作轴于点. ∵.在中... 在中... ∴. ∴．··············································· 6分 ②当.设交于点.作轴于点. .则四边形是等腰梯形. 重叠部分的面积是． ∴. ∴．··········· 8分 ③当.设与交于点.交于点.重叠部分的面积是． 因为和都是等腰直角三角形.所以重叠部分的面积是． ∵.. ∴. ∴. ∴ ．················································ 10分 (3)存在 ···································································································· 12分 ··································································································· 14分

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