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1问题的提出过抛物线y=x2上一点A(1,1)作抛物线的切线,分别交x轴于D,交y轴于B.点C在抛物线上,点E在线段AC上,满足EAEC=λ1;点F在线段BC上,满足FBCF=λ2,且λ1+λ2=1,线段CD与EF交于点P.当点C在抛物线上移动时,求点P的轨迹方程.2问题的解决解抛物线在点A处的切线斜率为y′=2x|x=
1 The problem is raised by a parabola y=x2 where the point A(1,1) is tangent to the parabola, intersecting the x axis with D, intersecting the y axis with B. point C on the parabola, and point E on the line segment AC, satisfying EAEC. =λ1; point F on line BC, satisfies FBCF=λ2, and λ1+λ2=1, line segment CD and EF intersect at point P. When point C moves on the parabola, the trajectory equation of point P is found. Solve the tangent slope of the solution parabola at point A as y’=2x|x=