设直线y=2x是抛物线y=x²+ax+b上过(2,4)处的切线
先求导:y‘=2x+a,因为直线的斜率为2,将点(2,4)带入求导式中:2=2*2+a,解得a=-2;将点(2,4)带入抛物线4=2*2+(-2)*2+b,解得:b=4.
绛旓細锛1锛夌敱棰樻剰锛歏1=蟺鈭2a(2x2)2dx锛4蟺5x5|2a锛4蟺5(32?a5)V2锛澫鈭2a20[a2?x2]dy=蟺鈭2a20[a2?y2]dy=2蟺a4-蟺a4=蟺a4锛2锛夆埖V1+V2锛4蟺5(32?a5)+蟺a4鈭(V1+V2)鈥诧紳?4蟺a4+4蟺a3锛4蟺a3(a?1)浠わ紙V1+V2锛夆=0锛屽緱锛歛=1鑰屽綋0锛渁锛1鏃讹紝锛圴1+V2锛夆0鈭...
绛旓細2x=x^2 x(x-2)=0 x=0 x=2 浜ょ偣鍧愭爣(0,0)(2,4)
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绛旓細y² = 2x锛y = x 瑙f柟绋嬪緱浜ょ偣锛(0锛0)锛(2锛2)鍦ㄥ尯闂磝鈭圼0锛2]涓婏紝y = 鈭(2x) > y = x 鎵浠ュ畾绉垎鈭(0鈫2) [鈭(2x) - x] dx = 鈭2 * (2/3)x^(3/2) - x²/2 |(0鈫2)= (2鈭2/3) * 2鈭2 - 2 = 8/3 - 2 = 2/3 ...
绛旓細瑙 鐢 瑙e緱 鎵姹傜殑闈㈢Н鏄敱鏇茬嚎x=y+4,x=y2鍙鐩寸嚎y=-2,y=4鎵鍥存垚锛屽鍥5-15锛屾晠鏈 S锛濄(y+4)- y2銆昫y�锛(锛4y锛)锛18�
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
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绛旓細y1*y2=-p^2 褰揕涓嶅瀭鐩磝杞存椂 璁y=k(x-p/2)浠e叆y^2=2px 寰 k^2x^2-(pk^2+2p)x+k^2p^2/4=0 闊﹁揪瀹氱悊 x1x2=k^2p^2/4/k^2=p^2/4 y1y2 =k^2(x1x2-(x1+x2)*p/2+p^2/4)=-p^2 x1x2,y1y2涓哄畾鍊 濡傛灉鎮ㄨ鍙垜鐨勫洖绛旓紝璇风偣鍑烩滈噰绾充负婊℃剰绛旀鈥,绁濆涔犺繘姝ワ紒
绛旓細鈭鎶涚墿绾縴2=2px锛屽嵆x锛漼22p锛屼笌鐩寸嚎y=x浜ょ偣涓簒=y=2p姝ゆ椂锛寈鈥诧紳yp|y锛2p锛2锛寈鈥筹紳1p浜庢槸锛屽湪鐐癸紙2p锛2p锛夊鐨勬洸鐜囦负锛欿锛潀x鈥硘(1+x鈥2)32锛155p鍥犺屾洸鐜囧崐寰勪负锛歊锛1K锛55p锛55鈭磒=1鈭存姏鐗╃嚎鏂圭▼涓簓2=2x锛屼氦鐐规槸锛2锛2锛夛紝浜ょ偣澶勭殑y'=12鈭村湪杩欑偣澶勭殑鍒囩嚎鏂圭▼鏄細y?2锛12...
绛旓細y = -(x-2)2 +4 = -x2 +4x 锛2锛墆 = -x2 +4x 涓 y=2x 鑱旂珛 -(x-2)2 +4 = 2x 鍙樺舰寰楋細x2 - 2x = 0 瑙e緱x=0鎴杧=2 鍧愭爣锛欰(0,0) B(2,4)锛3锛夋湁銆傗柍ABP鐨勯潰绉瓑浜庯細AB闀垮害脳P鍒癆B璺濈脳1/2 AB闀垮害涓嶅彉锛岄渶瑕佹壘鍒拌窛绂籄B鏈杩滅殑P鐐 鐢ㄥ鏁板仛锛氫护P鐐瑰湪鎶涚墿绾涓...