抛物线y^2=2px(p>0)上有A(x1,y1)B(x2,y2)C(x3,y3)三点,F是它的焦点若|AF|,|BF|,|CF|成等差数列,则()A.2 x2 = x1 + x3B.2 y2 = y1 + y3C.2 x3 = x1 + x2D.2 y3 = y1 + y2
问题描述:
抛物线y^2=2px(p>0)上有A(x1,y1)B(x2,y2)C(x3,y3)三点,F是它的焦点若|AF|,|BF|,|CF|成等差数列,则()
A.2 x2 = x1 + x3
B.2 y2 = y1 + y3
C.2 x3 = x1 + x2
D.2 y3 = y1 + y2
答
因为A(x1,y1)B(x2,y2)C(x3,y3)三点在抛物线上
且抛物线上的点到焦点的距离等于点到准线的距离
所以
|AF|=x1+ p/2 |BF|=x2+ p/2 |CF|=x3+ p/2
又因为 |AF| , |BF| , |CF| 成等差数列
所以 2*|BF|=|AF|+|CF|
2*(x2+ p/2)=(x1+ p/2)+(x3+ p/2)
2 * x2 + p=x1 + x3 + p
则 2 x2 =x1+x3
答案选A
答
根据抛物线的第二定义可知:|AF|=x1+p/2;|BF|=x2+p/2;|CF|=x3+p/2.
又|AF|,|BF|,|CF|成等差数列,可知|AF|+|CF|=2|BF|,即2 x2 = x1 + x3.
故应选A.