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有这样一个命题:椭圆 x~2/a~2+y~2/b~2=1(a>b>0)短轴为 AB,M 为椭圆上非 A、B 的点,MA、MB 与 x 轴交于点 E、F,则 OE·OF=a~2.此命题看似平凡,却“来头”不小,值得研究.推广1:把短轴 AB,长轴 CD 换成一般的共轭直径,得到如下性质:定理1 AB、CD 是椭圆 x~2/a~2+y~2/b~2=1(a>b>0)的共轭直径,M 为椭圆上非 A、B 的点,直线 MA、MB 分别交 CD 所在直线于点 E、F,则 E、F 在点 O 的同侧,且 OE·OF=OD~2(图1).证明:设 A(acos α,bsin α),则 B(-acos α,-bsin α),M(acos β,bsin β).由 AB、CD 共轭,知 k_(AB)·k_(CD)=-(b~2/a~2),又 k_(AB)=bsin α/acos α,
There is such a proposition: the ellipse x~2/a~2+y~2/b~2=1 (a>b>0) the minor axis is AB, M is the point on the ellipse not A, B, MA, MB and The x-axis intersects at points E and F, then OE·OF=a~2. This proposition appears to be trivial, but it’s not small, so it is worth studying. Popularization 1: Replace short axis AB and long axis CD The conjugate diameter is given by: Theorem 1 AB, CD is the conjugate diameter of the ellipse x~2/a~2+y~2/b~2=1(a>b>0), M is non-elliptical Point A, B, the straight line MA, MB, respectively, to the CD where the straight line at points E, F, then E, F on the same side of the point O, and OE · OF = OD ~ 2 (Figure 1). Proof: Let A ( Acos α, bsin α), then B (-acos α, -bsin α), M (acos β, bsin β). Conjugate by AB, CD, know k_ (AB) · k_ (CD) = - (b ~ 2/a~2), again k_(AB)=bsin α/acos α,