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)��=8xk��x��-(4k��+8)x+4k��=0x1+x2=(4k��+4)/k��x1x2=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