Question

直线l:y=kx+根号2与椭圆C:x^2/3+y^2=1交于不同的两点A.B,且向量OA乘向量OB=1,求k值

Answer 1

把直线y=kx+√2代入椭圆C得x^2/3+(kx+√2)^2=1
所以(k^2+1/3)x^2+2√2kx+1=0
设A(x1,kx1+√2),B=(x2,kx2+√2)
由韦达定理得x1+x2=-2√2k/(k^2+1/3),x1*x2=1/(k^2+1/3)

因为OA*OB=1
所以x1*x2+(kx1+√2)*(kx2+√2)
=(k^2+1)x1*x2+√2k(x1+x2)+2
=(k^2+1)/(k^2+1/3)+√2k*(-2√2k)/(k^2+1/3)+2
=1
化简得k^2=2/3
所以k=±√6/3

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