Question

已知抛物线方程y^2=8x,则过点(2,0)被抛物线所截得的长为8的直线方程?

Answer 1

设直线方程为y=k(x-2)当无k时,x=2,与抛物线所截得的长为8,满足条件当有k时 ,联立方程得：k(x-2)=8xkx-(4k+8)x+4k=0x1+x2=(4k+4)/kx1x2=4y1+y2=k(x1+x2-4)=4/ky1.y2=k(x1-2)(x2-2)=-8/k线长d=√(x1-x2)+(y1-y2)=√(x1+x2)+(y1+y2)-4x1+x2)-4(y1+y2)=√(4k+4)/(k)+16/k-16=88k^4-5k-2=0k=(±√89+5)/16∴y=(±√89+5)(x-2)/16