設A,B 是橢圓x^2/9+y^/4=1的長軸兩個端點，P,Q是垂直于AB弦的端點，則AP與BQ焦點的軌跡方程是____KEY x^2/9-y^2/4=1

熱心網友

x^2/9+y^2/4=1---A(-3,0);B(3,0)假設P(x0,y0);Q(x0,-y0)因為點P、Q都在橢圓上，所以x0^2/9+y0^2/4=1---y0^2=4/9*(9-x0^2)---y0^2/(9-x0^2)=4/9......(*)那么AP、BQ的方程分別是y/(x+3)=y0/(x0+3)......(1);y/(x-3)=-y0/(x0-3)......(2)(1)*(2):y^2/(x^2-9)=-y0^2/(x0^-9)=y0^2/(9-x0^2)由(*):y^2/(x^2-9)=4/9---9y^2=4(x^2-9)---x^2/9-y^2/4=1.此即為所求的軌跡方程。