Question

已知抛物线y^2=6x,经过点M（4,3）做一弦,这条弦恰好被点M平分,求这条弦所在的直线方程

Answer 1

y-3=k(x-4)

y=kx+(3-4k)

y^2=k^2x^2+2k(3-4k)x+(3-4k)^2=6x

k^2x^2-(8k^2-6k+6)x+(3-4k)^2=6x

中点横坐标是(x1+x2)/2

韦达定理x1+x2=(8k^2-6k+6)/k^2

所以(4k^2-3k+3)/k^2=4

k=1

所以x-y-1=0