Question

高二数学圆锥曲线、导数

1.过抛物线x^2=4y的焦点F作直线交抛物线于P1(x1,y1),P2(x2,y2)两点，若y1+y2=6,则P1P2的距离为？
2.已知函数f(x)=f'(π/2)sinx+cosx,则f(π/4)=_________

Answer 1

x^2=4y的焦点为(0,1),设过焦点(0,1)的直线为y=kx+1

则令kx+1=x^2/4,即x^2-4kx-4=0

由韦达定理得x1+x2=4k,x1x2=-4

y1=k(x1)+1,y2=k(x2)+2

所以y1+y2=k(x1+x2)+2=4k^2+2=6,所以k^2=1

所以|AB|=|x1-x2|*根号(k^2+1)=根号{(k^2+1)[(x1+x2)^2-4x1x2]}

=根号{2*[16k^2+16]}=根号{2*32}=8
f(x)=f'(π/2)sinx+cosx
f'(x)=f'(π/2)cosx-sinx
所以f'(π/2)=-1
f(π/4)=f'(π/2)sin(π/4)+cos(π/4)
=-sin(π/4)+cos(π/4)=0