如果以抛物线y^2=4x过焦点的弦为直径的圆截y轴所得的弦长为4,求该圆的方程
问题描述：
如果以抛物线y^2=4x过焦点的弦为直径的圆截y轴所得的弦长为4,求该圆的方程
表示画出了图之后,看了网上的答案还是不懂.不懂这一步,r2=22+（r-1）2 .
答
y^2=4x
F(1,0)
AB:y=k(x-1)
x=(y+k)/k
y^2=4x=4*(y+k)/k
ky^2-4y-4k=0
yA+yB=4/k,yA*yB=-4
(yA-yB)^2=(yA+yB)^2-4yA*yB=(4/k)^2+16=16(1+k^2)/k^2
AB^2=(1+1/k^2)*(yA-yB)^2=16(1+k^2)^2/k^4
r^2=AB^2/4=4(1+k^2)^2/k^4
(yA+yB)/2=2/k
(xA+xB)/2=(2+k^2)/k^2
[x-(2+k^2)/k^2]^2+(y-2/k)^2=4(1+k^2)^2/k^4
x=0
|y1-y2|=2√[(3k^2+4)/k^2]=4
k=±2
r^2=6.25
(xA+xB)/2=(2+k^2)/k^2=1.5
(yA+yB)/2=2/k=±1
(x-1.5)^2+(y±1)^2=6.25