过抛物线y^2=2px(p>0)的焦点F的直线与抛物线交于AB,AB在抛物线准线上的射影为A',B',求∠A'FB'
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![过抛物线y^2=2px(p>0)的焦点F的直线与抛物线交于AB,AB在抛物线准线上的射影为A',B',求∠A'FB'](/uploads/image/z/5227851-3-1.jpg?t=%E8%BF%87%E6%8A%9B%E7%89%A9%E7%BA%BFy%5E2%3D2px%28p%3E0%29%E7%9A%84%E7%84%A6%E7%82%B9F%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%B8%8E%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%BA%A4%E4%BA%8EAB%2CAB%E5%9C%A8%E6%8A%9B%E7%89%A9%E7%BA%BF%E5%87%86%E7%BA%BF%E4%B8%8A%E7%9A%84%E5%B0%84%E5%BD%B1%E4%B8%BAA%27%2CB%27%2C%E6%B1%82%E2%88%A0A%27FB%27)
过抛物线y^2=2px(p>0)的焦点F的直线与抛物线交于AB,AB在抛物线准线上的射影为A',B',求∠A'FB'
过抛物线y^2=2px(p>0)的焦点F的直线与抛物线交于AB,AB在抛物线准线上的射影为A',B',求∠A'FB'
过抛物线y^2=2px(p>0)的焦点F的直线与抛物线交于AB,AB在抛物线准线上的射影为A',B',求∠A'FB'
您好!
如图,有AF=AA',BF=BB',AA'∥OF∥BB'
所以∠A'FO=∠FA'A=(180°-∠A)/2=90°-∠A/2
∠B'FO=∠FB'B=(180°-∠B)/2=90°-∠B/2
又∠A+∠B=180
∠A'FB'=∠FA'A+∠FB'B=180°-(∠A+∠B)/2=90°
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