已知A,B是椭圆x²/a²+y²/b²=1(a>b>0)的左右顶点,P为椭圆上异于A,B的任意一点,直线PA和PB的斜率分别为kPA,kPB求证:kPAkPB=-b²/a²
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![已知A,B是椭圆x²/a²+y²/b²=1(a>b>0)的左右顶点,P为椭圆上异于A,B的任意一点,直线PA和PB的斜率分别为kPA,kPB求证:kPAkPB=-b²/a²](/uploads/image/z/6937344-0-4.jpg?t=%E5%B7%B2%E7%9F%A5A%2CB%E6%98%AF%E6%A4%AD%E5%9C%86x%26%23178%3B%EF%BC%8Fa%26%23178%3B%EF%BC%8By%26%23178%3B%EF%BC%8Fb%26%23178%3B%EF%BC%9D1%28a%EF%BC%9Eb%EF%BC%9E0%29%E7%9A%84%E5%B7%A6%E5%8F%B3%E9%A1%B6%E7%82%B9%2CP%E4%B8%BA%E6%A4%AD%E5%9C%86%E4%B8%8A%E5%BC%82%E4%BA%8EA%2CB%E7%9A%84%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E7%9B%B4%E7%BA%BFPA%E5%92%8CPB%E7%9A%84%E6%96%9C%E7%8E%87%E5%88%86%E5%88%AB%E4%B8%BAkPA%2CkPB%E6%B1%82%E8%AF%81%EF%BC%9AkPAkPB%EF%BC%9D%EF%BC%8Db%26%23178%3B%EF%BC%8Fa%26%23178%3B)
已知A,B是椭圆x²/a²+y²/b²=1(a>b>0)的左右顶点,P为椭圆上异于A,B的任意一点,直线PA和PB的斜率分别为kPA,kPB求证:kPAkPB=-b²/a²
已知A,B是椭圆x²/a²+y²/b²=1(a>b>0)的左右顶点,P为椭圆上异于A,B的任意一点,直线PA和PB的斜率分别为kPA,kPB求证:kPAkPB=-b²/a²
已知A,B是椭圆x²/a²+y²/b²=1(a>b>0)的左右顶点,P为椭圆上异于A,B的任意一点,直线PA和PB的斜率分别为kPA,kPB求证:kPAkPB=-b²/a²
证明:
椭圆:(x²/a²)+(y²/b²)=1 (a>b>0)
易知,A(-a,0),B(a,0)
可设P(acost,bsint).
由斜率公式,可得:
Kpa=(bsint)/(acost+a)
Kpb=(bsint)/(acost-a)
∴Kpa·Kpb
=[(bsint)/(acost+a)]×[(bsint)/(acost-a)]
=(b²/a²)[(sin²t)/(cos²t-1)]
=(b²/a²)[(sin²t)/(-sin²t)]
=-b²/a²
已知,A(-a,0)B(a,0)
设P(x,y)则kPA=y/(x+a)
kPB=y/(x-a)
故,kPAkPB=y2/(x2-a2)。。。。。。1
又,x²/a²+y²/b²=1。。。。。。。。2
由1,2两式化简得,kPAkPB=-b2/a2