Question

一道抛物线的题，麻烦各位帮忙啦……

过抛物线Y的平方=2px(p>0)的焦点F作直线交抛物线A,B两点,过A,B分别向抛物线的准线作垂线,垂足A1,B1,已知四边形AA1B1F与BB1A1F的面积分别为15和7,则三角形A1B1F的面积为?求详解

Answer 1

设A(x1,y1) B(x2,y2) 设y1>0 y2<0
过焦点F的直线 y=k(x-p/2) x=y/k+p/2
y^2=2px y^2=2p(y/k+p/2)
y^2-2py/k-p^2=0
AA1=x1+p/2 BB1=x2+p/2 A1B1=(y1-y2)
S=S四边形AA1BB1=(AA1+BB1)*A1B1/2=(x1+x2+p)*(y1-y2)/2
S1=S△AA1F=1/2*AA1*y1=(x1+p/2)*y1/2
S2=S△BB1F=1/2*BB1*|y2|=-(x2+p/2)*y2/2
所以S1+S2=S/2
所以S△A1B1F=S/2=S1+S2
S四边形AA1B1F=S△A1B1F+S1=15
S四边形BB1A1F=S△A1B1F+S2=7 相加
2S△A1B1F+S1+S2=22
3S△A1B1F=22
SA1B1F=22/3