已知抛物线y^2=2px(p>0),焦点为F,一直线l与抛物线交于A、B两点,且AF+BF=8,且AB的垂直平分线恒过定点S(6,0)(1)求抛物线方程(2)求三角形ABS面重点是面积?
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![已知抛物线y^2=2px(p>0),焦点为F,一直线l与抛物线交于A、B两点,且AF+BF=8,且AB的垂直平分线恒过定点S(6,0)(1)求抛物线方程(2)求三角形ABS面重点是面积?](/uploads/image/z/1807205-5-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%5E2%3D2px%28p%3E0%29%2C%E7%84%A6%E7%82%B9%E4%B8%BAF%2C%E4%B8%80%E7%9B%B4%E7%BA%BFl%E4%B8%8E%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%BA%A4%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%E4%B8%94AF%2BBF%3D8%2C%E4%B8%94AB%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BF%E6%81%92%E8%BF%87%E5%AE%9A%E7%82%B9S%EF%BC%886%2C0%EF%BC%89%281%29%E6%B1%82%E6%8A%9B%E7%89%A9%E7%BA%BF%E6%96%B9%E7%A8%8B%EF%BC%882%EF%BC%89%E6%B1%82%E4%B8%89%E8%A7%92%E5%BD%A2ABS%E9%9D%A2%E9%87%8D%E7%82%B9%E6%98%AF%E9%9D%A2%E7%A7%AF%3F)
已知抛物线y^2=2px(p>0),焦点为F,一直线l与抛物线交于A、B两点,且AF+BF=8,且AB的垂直平分线恒过定点S(6,0)(1)求抛物线方程(2)求三角形ABS面重点是面积?
已知抛物线y^2=2px(p>0),焦点为F,一直线l与抛物线交于A、B两点,且AF+BF=8,且AB的垂直平分线恒过定点S(6,0)
(1)求抛物线方程
(2)求三角形ABS面
重点是面积?
已知抛物线y^2=2px(p>0),焦点为F,一直线l与抛物线交于A、B两点,且AF+BF=8,且AB的垂直平分线恒过定点S(6,0)(1)求抛物线方程(2)求三角形ABS面重点是面积?
答:
① 焦点在x轴上,可设抛物线方程为:y² = 2px.可以判断焦点在(p/2,0)点.
② 设A点坐标(x1,y1),B点坐标(x2,y2),设AB斜率是k,线段AB的垂直平分线斜率是k'
则:kk' = -1,所以:
(y1-y2)/(x1-x2) * [(y1+y2)/2 - 0 ]/[(x1+x2)/2 - 6] = -1
(y1² - y2²) / [x1² - x2² -12(x1 - x2)] = -1
代入y1²=2px1,y2²=2px2,化简:
2p/(x1 + x2 - 12) = -1
x1 + x2 = 12 - 2p ---
③
AF²=(x1 - p/2)² + y1² = (x1 - p/2)² + 2px1 = (x1 + p/2)²
AF = x1 + p/2
同理:
BF = x2 + p/2
AF + BF = x1 + x2 + p ---
link:
12 - 2p + p = 8
p=4
综上:
抛物线方程:
y² = 8x