Question

抛物线x^2=-y,直线y=kx-1与抛物线交于A,B,AB绝对值=2√10,求k

Answer 1

代入,得：x^2+kx-1=0,所以|x1-x2|=√[(x1+x2)^2-4x1x2]=√(k^2+4)

|AB|=2√10,推得：√(k^2+1)√(k^2+4)=210,即得：(k^2+1)(k^2+4)=40

k^4+5k^2-36=0,解得：k^2=4(k^2=-9舍去）,k=±2