Question

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)...

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
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b^2X0^2/a^2,X1^2/a^2+y1^2/b^2=1,则y1^2=
b^2
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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.