A、B是椭圆b2x2+a2y2=a2b2（a>b>0）的左右顶点,P是椭圆上异于A、B一点,求证:kPA×kPB=-b2/a2

Question

A、B是椭圆b2x2+a2y2=a2b2（a>b>0）的左右顶点,P是椭圆上异于A、B一点,求证:kPA×kPB=-b2/a2

Answer 1

证明：椭圆b2x2 + a2y2 = a2b2（a > b > 0）,x2 /a2 + y2 /b2 = 1,左顶点A(-a,0),右顶点B(a,0),点P是椭圆上易于点A,B的一点,三角代换可设点P(acosθ,bsinθ),θ∈(0,π)∪(π,2π),所以kPA = (bsinθ)/(acosθ + a),kPB = (bsinθ)/(acosθ – a),所以kPA *kPB = (b2sin2θ)/(a2cos2θ – a2) = (b2/a2)(1 – cos2θ)(cos2θ – 1) = -b2/a2 ,证毕.