发布网友 发布时间:2024-10-23 20:56
共2个回答
热心网友 时间:2024-11-08 18:17
(1)
题中没提顶点,但看来应当是O
OP = √(3² + 3²) = 3√2
E(3√2, 0)
PE的斜率k = (3 - 0)/(3 - 3√2) = 1/(1 - √2)
PF的斜率 = -1/k = √2 - 1
PF的方程: y - 3 = (√2 - 1)(x - 3)
x = 0, y = 3(2 - √2)
F(0, 3(√2 - 2))
(2)
(i) t < 0 (图1, 极端的情况见红线)
EP的方程: y = 3(x - t)/(3 - t)
x =0, y = 3t/(t - 3), 与y轴交于Q(0, 3t/(t - 3))
其垂线的方程: y - 3 = (t - 3)(x - 3)/3
y = 4, x = 3(t - 2)/(t - 3), 与BC交于R(3(t-2)/(t - 3), 4)
S = 三角形PQW面积 + 梯形CRPW面积
= (1/2)*3*[3 - 3t/(t - 3)] + (1/2)*[3(t-2)/(t - 3) + 3](4 - 3)
= 3(t - 7)/(t - 3)
(ii) 0 ≤ t < 2 (图2, 极端的情况见天蓝线)
EP垂线的方程: y - 3 = (t - 3)(x - 3)/3
y = 4, x = 3(t - 2)/(t - 3), 与BC交于R(3(t-2)/(t - 3), 4)
S = 梯形CRPW面积 + 梯形OEPW面积
= (1/2)*[3(t-2)/(t - 3) + 3](4 - 3) + (1/2)(3 + t)*3
= (3/2)(t² + 2t - 14)/(t- 3)
(iii) 0 ≤ t < 3 (图3, 极端的情况见紫线)
EP垂线的方程: y - 3 = (t - 3)(x - 3)/3
x = 0, y = 6 - t, 与y轴交于R(0, 6 - t)
显然三角形RPW与三角形EPS全等
S = 正方形OSPW的面积
= 9
(iv) 3 ≤ t < 4(图4, 极端的情况见绿线)
显然三角形RPW与三角形EPS全等
S = 正方形OSPW的面积
= 9
(v) 4 ≤ t < 6(图5, 极端的情况见粉红线)
PE与AB交于Q(4, 3(4 - t)/(3-t))
EP垂线与y轴交于R(0, 6 - t)
S = 梯形ORPS面积 + 梯形SAQP面积
= (1/2)*3(3 + 6 - t) + [(1/2)*1*(3 + 3(4 - t)/(3 - t)]
= (3/2)(t² - 14t + 34)/(3 - t)
其余情况可类似做。有时间再补充。
热心网友 时间:2024-11-08 18:19
第一问:应该分情况来讨论
(1)当PO=OE时 F(0,0), F(0,6+3√2 )
(2)当EP=EO时 F(0,3),
(3)当PE=PO时 F(0,6-3√2 ),