x2−1 |
∵∠P=∠P,∠PCD=∠A=90°,
∴Rt△PCD∽Rt△PAB,
∴
AB |
CD |
PA |
PC |
∴AB=
CD•PA |
PC |
x+1 | ||
|
设y=AB•S△PAB,代入可得y=
(x+1)3 |
2(x2−1) |
(x+1)2 |
2(x−1) |
去分母,得x2+2(1-y)x+1+2y=0,
因为x是实数,所以△=4(1-y)2-4(1+2y)=4y(y-4)≥0,
又因为y>0,所以y≥4.即y的最小值为4,故当PD=3时,AB•S△PAB的最小值为4.
答:AB•S△PAB的最小值是4.
x2−1 |
AB |
CD |
PA |
PC |
CD•PA |
PC |
x+1 | ||
|
(x+1)3 |
2(x2−1) |
(x+1)2 |
2(x−1) |