Question

求证：抛物线y^2=2x+ p

Answer 1

证明：设抛物线为y^2=2px(p>0),过焦点f(p/2,0)的弦直线方程为y=k(x-p/2),直线与抛物线交于a（x1,y1),b(x2,y2)
联立方程得k^2(x-p/2)^2=2px,整理得k^2x^2-p(k^2+2)x+k^2p^2/4=0
所以x1+x2=p(k^2+2)/k^2
由抛物线定义，af=a到准线x=-p/2的距离=x1+p/2,
bf=x2+p/2
所以ab=x1+x2+p=p(1+2/k^2+1)=2p(1+1/k^2)=2p(1+cos^2/sin^2a)=2p/sin^2a
证毕