Question

已知抛物线c：x^2=2py(p>0)的焦点为f，抛物线上一点a的横坐标为x1（x1>0），过点

Answer 1

由条件可以求得：
抛物线方程
y=x^2/4，p=2，L:y=0.5，右
切点
为(2√3,3);
Spmn=0.5|mn|*h,
其中h=p/2-y(p)=1-y(p)
设切点(x1,x1^2/4)，(x2,x2^2/4)
=>
切线：
y-x1^2/4=0.5x1(x-x1)=0.5x1x-0.5x1^2
y-x2^2/4=0.5x2(x-x2)=0.5x2x-0.5x2^2
也即：
x2y+x2x1^2/4=0.5x1x2x
x1y+x1x2^2/4=0.5x1x2x
消去x，求P点
纵坐标
=>
(x2-x1)y=x1x2(x2-x1)/4
=>
y=0.25x1x2=p(y)
=>
h=p/2-p(y)
=1-0.25x1x2
下面求|MN|
令y=p/2=1
=>
1+x1^2/4=0.5x1xm
1+x2^2/4=0.5x2xn
=>
x2+x2x1^2/4=0.5x1x2xm
x1+x1x2^2/4=0.5x1x2xn
=>
(xm-xn)=(x2-x1)(1-0.25x1x2)/(0.5x1x2)
=>
|MN|=|x2-x1|(1-0.25x1x2)/(0.5x1x2)
=>
Spmn=0.5|mn|*h
=0.5(1-0.25x1x2)|x2-x1|(1-0.25x1x2)/(0.5x1x2)
=(1/x1-1/x2)(0.25x1x2-1)^2
S(x2)=(1/x1-1/x2)(0.25x1x2-1)^2
S'(x2)=(1/x1-1/x2)*2(0.25x1x2-1)*0.25x1+(0.25x1x2-1)^2*(1/x2^2)
令S'(x2)=0
=>
2x2^2-x1x2-4=0
代入x1=2√3得到x2=0.5(√3-√11)

Answer 2

焦点(0,
p/2)
y₁
=
(x₁)²/(2p),
y₂
=
(x₂)²/(2p),
y₁y₂
=
(x₁x₂)²/(4p²)
=
p²/4
x₁x₂
=
-p²
(按图，a,
b在y轴两侧)
ab的方程:
[y
-
(x₂)²/(2p)]/[
(x₁)²/(2p)
-
(x₂)²/(2p)]
=
(x
-
x₂)/(x₁
-
x₂)
2py
=
(x₁
+
x₂)x
-
x₁x₂
=
(x₁
+
x₂)x
+
p²
x=0,
y
=
p/2,
ab过抛物线c的焦点