Question

过抛物线y=ax^2（a>0)的焦点F做一直线交抛物线与P ,Q两点，若线段PF与PQ的长分别是p.q则，(1/p)+(1/q)等于

答案用a表示，上面的PQ改成FQ

Answer 1

显然焦点为：（1/4a,0)
准线为y=-1/4a
设直线PQ为y=k(x-1/4a),P(x1,y1),Q(x2,y2)
将直线代入抛物线方程消去x
a(y/k+1/4a)²-y=0
ay²/k²+y(1/2k-1)+1/16a=0
y1+y2=-(1/2k-1)/a,y1*y2=1/16a²
由于抛物线上的点到焦点的距离等于到准线的距离
则p=y1+1/4a,q=y2+1/4a
则1/p+1/q=(y1+y2+1/2a)/[y1*y2+(y1+y2)/4a+1/16a²]
=(1-1/2k+1/2)/a/[1/16a²-(1/2k-1)/4a²+1/16a²]
=(3-k)/2a/[(3-k)/8a²]=4a

Answer 2

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