直线与圆锥曲线已知直线l：y=x+1与焦点在x轴上的椭圆x^2/5+y^2/m=1,相交于P、Q两点,且OP垂直OQ,其中O为坐标原点,求m的值.

Answer 1

设P(x1,y1),Q(x2,y2)
y=x+1
x^2/5+y^2/m=1
mx^2+5y^2=5m
mx^2+5(x+1)^2=5m
(m+5)x^2+10x+5-5m=0
x1+x2=-10/(m+5),
x1x2=(5-5m)/(m+5)
由OP⊥OQ,得
x1x2+y1y2=0,
即x1x2+(x1+1)(x2+1)=0
(x1+x2)+2x1x2+1=0
-10/(m+5)+2*(5-5m)/(m+5) +1=0
-10+2*(5-5m)+m+5 =0
-10+10-10m+m+5 =0
-10m+m+5 =0
9m=5
m=5/9