1、5T阿基米德三角形的几个结论熊昌进( +E 616650)孔繁秋(0 361009)李迪淼( 2 =v 410006)I:1 L 1, 0)B.5T L S= 1p (y20 - 2px0 ) 3 ( = 1p f (x0 , y0 ) 32 ) ,f (x0 , y0 )= y20 - 2px0 .2. T Ly2 = 2px (p 0)B,T L S,5TEB L:y2= 2px+ 3 S2p2 .3. M(m, n) Ly2= 2px ( p 0) =B,5VM L?H Kl1p ( 2pm- n2 )3 . ( = 1p - f (m, n) 32 )f (m,n)= n2- 2pm
2、0) ,l BT L HAVM( ca , - pba ) , OTMx H,N |Kl1a3 p( 2ac- pb2 ) 3.=、1=Q wL .=Q wL( C): ( 1- l2 )x2+ y2 - 2px+ p2 = 0( p 0) , M(m,n) wL( C) O .VM T wL( C)?AB,5 wL( C)VA, B MLT (x0 , y0 ) ,5TAB wL( C)MM , T , AB H.1.VM?ABH wL( C) EBL, Z:( 1- l2 )m- p x+ ny+ p2 - pm= 0.w 1: MM(xm , 0) ,5TExL Lx= p2 - pmp
3、- m( 1- l2 ) .w 2: M(p, 0) ,5TEMMLx= 0.w 3:MM wL( C) TEM? .M?AB, O !A (x1 , y1 ) , B( x2 , y2 ) ,5|( 1- l2 )x21+ y21 - 2px1+ p2= 0( 1-l2 )x22+ y22 - 2px2+ p2= 0 TMh, VAB| qTE(L)| q(+ y H, 90)M.2. Ll: Ax+ By+ C= 0 wLC: ( 1- l2 )x2+ y2 - 2px+ p2 = 0 ,5l BT wLC HAV. N1 5BB4. - ,1 +Yi,BB . )(武汉市二中田化澜整理)271999 M10 Y
