lucy3882387
幼苗
共回答了14个问题采纳率:100% 举报
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如果不懂,请Hi我,祝学习愉快!
1年前
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