Question

过抛物线y =ax^2(a>0)的焦点F作一直线交抛物线于P、Q两点,若线段PF与FQ的长

Answer 1

抛物线标准方程：x^2=y/a

F(0,1/(4a)),设P(x1,y1) Q(x2,y2),

PQ平行于X轴时,方程为：y=1/(4a),p=q=1/(2a),1/p+1/q=4a

PQ不平行于X轴时,设其方程为x=k(y-1/(4a))

代入抛物线方程得：y=ak^2*(y-1/(4a))^2

16ay=k^2(4ay-1)^2

16(aky)^2-(8ak^2+16a)y+k^2=0

y1+y2=1/(2a) + 1/(ak^2)

y1y2=1/(16a^2)

准线：y=-1/(4a)

PF=y1+1/(4a),FQ=y2+1/(4a)

p+q=y1+y2+1/(2a)=(k^2+1)/(ak^2)

pq=y1y2+(y1+y2)/(4a)+1/(4a)^2

=1/(16a^2) + 1/(8a^2) + 1/(4a^2k^2) + 1/(16a^2)

=(k^2+1)/(4a^2k^2)

1/q+1/p=(p+q)/(pq)=4a

综上可知,等于4a.

C