Question

知点P(3,2)是椭圆是x^2/25+y^2/16=1内一点(3)求过点P的直线被椭圆截的弦的中点Q的轨迹方程

Answer 1

弦AB的中点Q(x,y)

xA+xB=2x,yA+yB=2y

k(AB)=k(PQ)

(yA-yB)/(xA-xB)=(y-2)/(x-3)

[(xA)^2/25+(yA)^2/16]-[(xB)^2/25+(yB)^2/16]=1-1

16*(xA+xB)*(xA-xB)+25*(yA+yB)*(yA-yB)=0

16*2x+25*2y*(yA-yB)/(xA-xB)=0

16x+25y*(y-2)/(x-3)=0

(x-1.5)^2/(61/16)+(y-1)^2/(61/25)=1