求解这道高二双曲线数学题，求详解，为什么是选B，O(∩
\u9ad8\u4e2d\u6570\u5b66\uff0c\u53cc\u66f2\u7ebf\u95ee\u9898\u3002 \u4e3a\u4ec0\u4e48\u9009b\uff1f\u56e0\u4e3aa\u65b9=2\uff0c\u6240\u4ee5c\u65b9>1\uff0c\u6240\u4ee5\u76f4\u89d2\u4e09\u89d2\u5f62\u76f4\u89d2\u9876\u70b9\u53ea\u80fd\u591f\u5728\uff081,\u6839\u53f72\uff09\u3002
\u8bb0\u5f97\u76f4\u89d2\u4e09\u89d2\u5f62\u4e2d\u7684\u5c04\u5f71\u5b9a\u7406\u5417\u3002
\u8fc7\u76f4\u89d2\u9876\u70b9\u505a\u659c\u8fb9\u5782\u7ebf\uff0c\u5782\u8db3\u628a\u659c\u8fb9\u5206\u4e3a\u4e24\u4e2a\u90e8\u5206\uff08c-1\uff09\uff0c\uff08c+1\uff09\uff0c\u9ad8\u6839\u53f72\uff0c\u6240\u4ee5\uff08c-1\uff09*\uff08c+1\uff09=\uff08\u6839\u53f72\uff09\u65b9\u3002\u89e3\u5f97c\u65b9=3\uff0c\u6240\u4ee5b=1\u3002
\u9996\u5148\uff0c\u65e0\u8bba\u53cc\u66f2\u7ebf\u79bb\u5fc3\u7387\u591a\u5927\uff0cOA*OB \u90fd\u6ca1\u6709\u6700\u5927\u503c\u3002
\u8fd9\u662f\u7531\u4e8e A\u3001B \u53ef\u4ee5\u8fdc\u79bb\u9876\u70b9\u5230\u65e0\u7a77\uff0c|OA|\u3001|OB| \u53ef\u4ee5\u8fbe\u65e0\u7a77\uff0c
\u800c\u89d2 AOB \u8d8b\u4e8e 0 \u5ea6\uff08A\u3001B \u8d8b\u4e8e\u91cd\u5408\u81f3\u65e0\u7a77\u8fdc\uff09\uff0ccosAOB \u8d8b\u4e8e 1 \u3002
\u8fd9\u6837\u5c31\u80fd\u6392\u9664 A\u3001B \u3002
\u5f53\u53cc\u66f2\u7ebf\u79bb\u5fc3\u7387\u5927\u4e8e \u221a2 \u65f6\uff0c\u53cc\u66f2\u7ebf\u7684\u6e10\u8fd1\u7ebf\u5728 x \u8f74\u6b63\u65b9\u5411\u7684\u5f20\u89d2\u4e3a\u949d\u89d2\uff0c
\u56e0\u6b64\u5f53 A\u3001B\u4e00\u4e2a\u5728\u4e0a\u534a\u652f\uff0c\u4e00\u4e2a\u5728\u4e0b\u534a\u652f\u5e76\u4e14\u8fdc\u79bb\u9876\u70b9\u65f6\uff0c
\u53ef\u4ee5\u4f7f\u89d2 AOB \u4e3a\u949d\u89d2\uff08\u8fd9\u4e2a\u949d\u89d2\u9010\u6e10\u8d8b\u4e8e\u4e00\u4e2a\u5b9a\u503c\uff09\uff0c\u4e14 OA*OB < 0 \uff0c\u8d8b\u4e8e\u8d1f\u65e0\u7a77\uff0c
\u6240\u4ee5\u65e0\u6700\u5c0f\u503c\u3002
\u9009 D \u3002
解法如下：
圆与双曲线右支的切点记为Q
MO=PQ/2 MT=PF/2-FT
显然OF=c，OT=a，所以FT=b
所以得到：MO-MT=PQ/2-(PF/2-b)=(PQ-PF)/2+b=b-a
绛旓細杩欓亾棰涓嶉毦銆傝В娉曞涓嬶細鍦嗕笌鍙屾洸绾鍙虫敮鐨勫垏鐐硅涓篞 MO=PQ/2 MT=PF/2-FT 鏄剧劧OF=c锛孫T=a锛屾墍浠T=b 鎵浠ュ緱鍒帮細MO-MT=PQ/2-(PF/2-b)=(PQ-PF)/2+b=b-a
绛旓細鐢遍鎰忓緱c=6锛屾晠a^2+b^2=c^2=36---(1)鍙堝浘鍍忚繃(2,-5)锛屼唬鍏ュ緱25/a^2-4/b^2=1---(2)(1)(2)鑱旂珛娑堝幓b寰楀叧浜巃^2鐨勬柟绋嬶紝a^4-65a^2+900=0锛岃В寰梐^2=20鎴45锛屽張c^2=36锛宎<c锛屾墍浠^2=20锛屼唬鍏ヤ换涓鏂圭▼姹傚緱b^2=16銆傛晠姝鍙屾洸绾鏂圭▼涓簓^2/20-x^2/16=1 ...
绛旓細璁惧弻鏇茬嚎鐨勭劍鐐瑰湪x杞翠笂锛屼笖杩囩偣A(1, 0)鍜孊(-1, 0)锛孭鏄弻鏇茬嚎涓婂紓浜嶢B鐨勪换鎰忎竴鐐癸紝濡傛灉涓夎褰PB鐨勫瀭蹇僅鎬诲湪姝ゅ弻鏇茬嚎涓锛屾眰鍙屾洸绾鐨勬爣鍑嗘柟绋嬨傝В锛氳P(x0, y0)銆傚洜涓篐涓哄瀭蹇冿紝浜庢槸PH鈯B銆丄H鈯P銆侫B鍦▁杞翠笂锛屾墍浠H鍨傜洿浜巟杞达紝PH鎵鍦ㄧ洿绾夸笌鍙屾洸绾垮彧鏈変袱涓氦鐐癸紝涓涓槸P(x0, y0...
绛旓細F1(鈭5,0)锛屼腑鐐逛负(0,2)锛屾墍浠鐐瑰潗鏍囦负(鈭5,4)浠ｅ叆鍙屾洸绾鏂圭▼鍙互寰楀埌 a^2=1锛宐^2=4 鎵浠ュ弻鏇茬嚎鏂圭▼涓簒^2-y^2/4=1
绛旓細=36k^4+4(72k²-9k^4+64-8k²)=16鈭(k²+1)x1,2=[6k²卤16鈭氾紙k²+1锛塢/2(8-k²)锝淏F2锝-锝淏F1锝=锝淎F1锝-锝淎F2锝=2a=2 鈭达綔BF2锝-锝淎F2锝=锝淎B锝=4 锝淎B锝²=(x1-x2锛²+锛坹1-y2锛²=锛1+k²锛夛紙x1-x2)...
绛旓細鍙屾洸绾鐒﹁窛2c涓10銆2銆佸弻鏇茬嚎绂诲績鐜噀=5/3, 瀹炶酱闀2a=6,a=3,e=c/a=5/3,c=5a/3=5,铏氬崐杞碽=鈭氾紙c^2-a^2)=4,铏氳酱鐨勯暱涓8锛岀劍璺濅负2c=10.3銆佺劍璺2c=8,c=4,铏氬崐杞碽=6/2=3,瀹炲崐杞碼=鈭氾紙c^2-b^2)=鈭7,瀹炶酱闀夸负2鈭7,绂诲績鐜噀=c/a=4/鈭7=4鈭7/7....
绛旓細a=卤m,b=卤2m c=卤鈭5 |m| 娓愯繎绾挎柟绋媦=卤b/a x y=卤2x 2 鐩寸嚎l鏂滅巼涓1杩囧彸鐒︾偣F y=x+b b=-c= -鈭5|m| y=x-c浜鍙屾洸绾鍒嗗埆浜や簬P1(x1,y1)锛孭2(x2,y2)涓ょ偣 4x^2-(x-c)^2=4m^2 3x^2+2cx-4m^2-c^2=0 3x^2+2cx-9m^2=0 x1+x2=-2c/3 x1x2=-9m^2...
绛旓細鍙屾洸绾鐨勪竴涓《鐐圭殑鍧愭爣涓(0,2)锛屽垯鍙煡鍙屾洸绾跨劍鐐瑰湪y杞翠笂锛屼笖a=2锛屽疄杞撮暱涓庤櫄杞撮暱涔嬪拰绛変簬鍏剁劍璺濈殑鈭2鍊,鍒2a+2b=鈭2•2c 涓旀湁c²=a²+b²,鍗2+b=鈭2 c,c²=4+b²,瑙ｅ緱b=2,c=2鈭2.鈭村弻鏇茬嚎鏍囧噯鏂圭▼涓簓²/4-x²/4=1....
绛旓細|PF₂|²=|F₁F₂|²=锔盤F₁锔²+|F₁F₂|²-2锔盤F₁锔眧F₁F₂锔眂os鈭燩F₁F₂鏁呭緱锔盤F₁锔-(16c/5)=0锛屽嵆鏈夛副PF₁锔=16c/5.锔盤F₁锔-|PF₂|=锔盤F₁锔...
绛旓細渚濇嵁棰樼洰鎰忔濓紝鍙嶆帹鍙屾洸绾鏂圭▼锛歈鐐瑰潗鏍(c/2,鈭3c/2)锛孎鐐瑰潗鏍(-c,0)锛孧鐐瑰潗鏍(c,0)銆俀F-QM=2a锛屽甫鍏ュ潗鏍囷紝a=(鈭3-1)/2锛宎²=(2-鈭3)/2銆傚張a²锛媌²=c²锛宐²=鈭3/2銆傚弻鏇茬嚎鏂圭▼涓簒²/[(2-鈭3)/2]-y²/(鈭3/2)=1 ...
