Question

高中数学椭圆，第九题。谢谢

Answer 1

原点为O
考虑四边形OAPB,
若PA与PB垂直, 则四边形OAPB为正方形.
OP = 2^0.5 * b (1)
只要P点坐标满足(1)式, 那么, 必然有PA与PB垂直
设P坐标为(a cos(s), b sin(s))
(a cos(s))^2 + (b sin(s))^2 = (2^0.5 * b)^2
cos(s) ^2 = b^2 / (a^2 - b^2) = 1 / e^2 - 1 (2)
s 有解, 则 -1 <= cos(s) <= 1
0 <= 1 / e^2 - 1 <= 1
e >= 2^0.5 / 2
所以,
e < 2^0.5 / 2时, P不存在
e >= 2^0.5 / 2时, 存在P满足PA与PB垂直. 此时
由(2)式可得
s = arccos((1 / e^2 - 1)^0.5)
cos(s) =± (1 / e^2 - 1)^0.5
sin(s) = ± (2 - 1 / e^2)^0.5
则P坐标为(±a (1 / e^2 - 1)^0.5, ±b (2 - 1 / e^2)^0.5)