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X^2/a^2+y^2/b^2=1,
c/a=√2/2,a=√2c,a^2=2c^2,b^2=a^2-c^2=2c^2-c^2=c^2,b=c
(1,√2/2)代入1/a^2+1/(2b^2)=1,1/( 2c^2)+1/(2c^2)=1,c=1=b,a=√2
椭圆为x^2/2+y^2=1 ,右焦点为(1,0)
设直线为y=k(x-1),x=ty+1,(t=1/k) ,代入x^2/2+y^2=1,(ty+1)^2+2y^2-2=0
(t^2+2)y^2+2ty-1=0 ,⊿=8t^2+8>0,所以有二个不等实根,
y1+y2=-2t/(t^2+2) y1y2=-1/(t^2+2)
|y1-y2|^2=(y1+y2)^2-4y1y2
=[-2t/(t^2+2)]^2-4(-1)/(t^2+2)=(4t^2+4t^2+8)/(t^2+2)^2=8(t^2+1)/(t^2+2)^2
(2√6/7)^2=[1/2*1*|y1-y2|]^2
24/49=1/4*8(t^2+1)/(t^2+2)^2
12t^4-t^2-1=0,t^2=1/3,k^2=3 ,k= ±√3
直线为y=±√3(x-1)
1年前
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