Question

如图，设抛物线C：y^2=2px(p＞0)的焦点为F，过点F的直线l1交抛物线C于A,B两点，且|AB|=8，

线段AB的中点到y轴的距离为3.
（1）求抛物线C的方程
（2）若直线l2与圆x^2 +y^2=1/2切于点P，与抛物线C切于点Q，求△FPQ的面积。求详解，要过程。谢谢！

Answer 1

(1)
F(0.5p,0),p>0
y=k(x-0.5p)
x=(y+0.5pk)/k
y^2=2px=2p*(y+0.5pk)/k
ky^2-2py-kp^2=0
yA+yB=2p/k,yA*yB=-p^2
(yA+yB)/2=p/k
(xA+xB)/2=(p+0.pk^2)/k^2=3
p+0.5pk^2=3k^2
p=(3-0.5p)k^2......(1)
(yA-yB)^2=(yA+yB)^2-4yA*yB=4p^2(1+k^2)/k^2
(xA-xB)^2+(yA-yB)^2=AB^2
(1+1/k^2)*4p^2(1+k^2)/k^2=8^2
p(1+k^2)=4k^2
p=(4-p)k^2......(2)
(1),(2):
p=2
C:y^2=4x
F(1,0)
(2)
x=my+b
y^2=4x=4(my+b)
y^2-4my-4b=0
(-4m)^2-4*(-4b)=0
b=-m^2
x=my-m^2
x-my+m^2=0
d^2=r^2=1/2
|0-0+m^2|^2/(1+m^2)=1/2
m^2=1
m=1,b=-1
L2:x-y+1=0
Q(1,2)
P( )
|PQ|=
h=d=|1-0+1|/√2=√2
s=(1/2)*|PQ|*h=