Question

跪求高手，已知椭圆标准方程椭圆的标准方程是x^2/a^2+y^2/b^2=1(a>b>0)

M、N是椭圆上关于原点对称的两点，P是椭圆上任意一点，且直线PM,PN的斜率分别为K1，K2，若K1K2的绝对值是1/4，求椭圆的离心率？

Answer 1

设P(x0,y0),M(x1,y1),N(x2,y2),
(1)式：X0^2/a^2+y0^2/b^2=1,
(2)式：X1^2/a^2+y1^2/b^2=1,
(3)式：X2^2/a^2+y2^2/b^2=1.
(2)-(1)得（X1^2-x0^2)/a^2+（y1^2-y0^2)/b^2=0,
整理得PM的斜率K1=(y0-y1)/(x0-x1)=-b^2(x0+x1)/a^2(y0+y1),
同理PN的斜率K2=(y0-y2)/(x0-x2)=-b^2(x0+x2)/a^2(y0+y2),
|K1*K2|=|[b^4(x0+x1)(x0+x2)]/[a^4(y0+y2)(y0+y1)]|=1/4,
M、N是椭圆上关于原点对称的两点,x1=-x2,y1=-y2.
[b^4(x0+x1)(x0+x2)]/[a^4(y0+y2)(y0+y1)]
=[b^4(x0+x1)(x0-x1)]/[a^4(y0-y1)(y0+y1)]
=[b^4(x0²-x1²)]/[a^4(y0²-y1²)]

因为P(x0,y0),M(x1,y1)在椭圆上，
所以X0^2/a^2+y0^2/b^2=1,则y0^2= b^2 - b^2X0^2/a^2，
X1^2/a^2+y1^2/b^2=1, 则y1^2= b^2 - b^2X1^2/a^2，
∴y0²-y1²=- b^2X0^2/a^2+ b^2X1^2/a^2
= b^2(X1^2 -X0^2)/a^2,
所以(x0²-x1²)]/(y0²-y1²)=-a^2/b2,
从而b^4(x0²-x1²)]/[a^4(y0²-y1²)=-b^2/a^2,
∵|K1*K2|=1/4,∴b^2/a^2=1/4,
(a^2-c^2)/a^2=1/4,
1- c^2/a^2=1/4,c^2/a^2=3/4,
∴e=√3/2.

【另法】
设P(acosβ，bsinβ),M(acosα，bsinα),则N(-acosα，-bsinα).
可得斜率k1= [b(sinβ-sinα)]/[a(cosβ-cosα)],
斜率k2= [b(sinβ+sinα)]/[a(cosβ+cosα)],
所以|k1*k2|=|[b²(sin²β-sin²α)]/[a²(cos²β-cos²α)]|
=|[b²((1-cos²β)-(1-cos²α))]/[a²(cos²β-cos²α)]|
=|[b²(-cos²β)+cos²α)]/[a²(cos²β-cos²α)]|
= b²/a²,
∵|K1*K2|=1/4,∴b^2/a^2=1/4,
(a^2-c^2)/a^2=1/4,
1- c^2/a^2=1/4,c^2/a^2=3/4,
∴e=√3/2.

Answer 2

设M(m,n)、N(-m,-n)、P(x,y)，则K1K2=(y-n)/(x-m)*(y+n)/(x+m)=(y^2-n^2)/(x^2-m^2)=1/4
∵M、N在椭圆上，∴m^2/a^2+n^2/b^2=1 (1)
∵P在椭圆上，∴x^2/a^2+y^2/b^2=1 (2)
(2)-(1)得(y^2-n^2)/b^2+(x^2-m^2)/a^2=0
(y^2-n^2)/(x^2-m^2)=-b^2/a^2
即b^2/a^2=1/4
b/a=1/2 e=sqr(3)/2