已知a，b，c為某一直角三角形的三邊長，c為斜邊，若點（m，n）在直線ax+by+2c=0上，則m2+n2的最小值為多少？需具體步驟。

熱心網友

ma+nb+2c=0-m=(b/a)*n + 2c/am^2 = (b^2/a^2)*n^2 + (4bc/a^2)*n + 4c^2/a^2TT = m^2+n^2 =(c^2/a^2)*n^2 + (4bc/a^2)*n + 4c^2/a^2=(c^2/a^2) * [ n^2 + (4b/c)*n + 4] TTT = n^2 + (4b/c)*n + 4= n^2 + 2*(2b/c)*n + (2b/c)^2 - (2b/c)^2 + 4= (n+2b/c)^2 + 4*a^2/c^2當 n=-2b/c 時 TTT 取最小值 4*a^2/c^2TT最小值 即為c^2/a^2 * 4 a^2/c^2 =4