解ç:æ ¹æ®é¢æ: ç´çº¿L:y=k(x-4);æç©çº¿:y^2=4x; (Kâ 0) èç«ä¸¤å¼åï¼æ´çå¯å¾: k^2X^2-(8k^2+4)x+16K^2=0; æ ¹æ®é¦è¾¾å®ç:X1+X2=8+k^2/4;X1X2=16; æ以:y1+y2=k(x1-4)+k(x2-4)=K(X1+X2)-8K=4/k;(Kâ 0) å æ¤:APçä¸ç¹o(X1/2+2;y1/2)为åå¿; åå¾R=|AP|/2=]1/2â[(X1-4)^2+y1^2] ; åç´çç´çº¿X=m; éè¿å¼¦é¿å
³ç³»å¯ä»¥ç¡®å®L: (L/2)^2+(m-X1)^2=R^2;æ ¹æ®é¢ç®å¯ä»¥ç¥é弦é¿è½ä¿æå®å¼ï¼ä¸ºäºè®¡ç®ä¸çæ¹ä¾¿å¯ä»¥ç¨ç¹æ®å¼æ³ãå³:åå®K=1; åæ:L^2/4=R^2-(m-X1)^2为ä¸ä¸ªå®å¼; L^2/4=12-4â5-20-4â5(m-6)-(m-6)^2; è¿ä¸æ¥æ´ç:å³è¾¹=-m^2-(4â5-12)m+28+20â5; æé å½æ°:F(X)=-X^2-(4â5-12)X+28+20â5;æ±å¯¼å¹¶ä»¤å¯¼æ°ä¸º0;åæ: -2X-4â5+12=0;解å¾X=6-2â5=X1å¼; æ
æ¤æ:å½M=6-2â5;满足ãä¹å°±æ¯è¯´åç´ç´çº¿X=6-2â5=XAå¼ã 1ï¼y1=a(x-k)^2+2(kï¼0),y1+y2=x^2+6x+12 =>y2=x^2+6x+12-y1 =>y2=x^2+6x+12-[a(x-k)^2+2]==>å½x=kæ¶ï¼y2=17 =>k^2+6k+12-2=17 ==>k1=1,k2=-7 ==>k>0==>k=1 2ï¼y2=x^2+6x+12-[a(x-k)^2+2] ==>y2=x^2+6x+12-[a(x-1)^2+2] ==>y2=[1-a]x^2+[6+2a]x+10-a ==>-b/2a=-[6+2a]/2[1-a]=-1 ==>a=-1 ==>y1=a(x-k)^2=-(x-1)^2=-x^2+2x-1 y2=[1+1]x^2+[6-2]x+10+1=2x^2+4x+11 3ï¼y1=y2==>-x^2+2x-1=2x^2+4x+11 ==>3x^2+2x+12=0==>Î=-140æ 交ç¹
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