已知圆C(x-3)²+(y-4)²=4,直线l1过定点A(1,0)若l1与圆相交于p、q两点,线段PQ的中点为M,又l1与l2:x+2y+2=0的交点为N,求证:|AM|*|AN|为定值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 20:51:44
![已知圆C(x-3)²+(y-4)²=4,直线l1过定点A(1,0)若l1与圆相交于p、q两点,线段PQ的中点为M,又l1与l2:x+2y+2=0的交点为N,求证:|AM|*|AN|为定值](/uploads/image/z/8628531-51-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%EF%BC%88x-3%29%26%23178%3B%2B%EF%BC%88y-4%29%26%23178%3B%3D4%2C%E7%9B%B4%E7%BA%BFl1%E8%BF%87%E5%AE%9A%E7%82%B9A%EF%BC%881%2C0%EF%BC%89%E8%8B%A5l1%E4%B8%8E%E5%9C%86%E7%9B%B8%E4%BA%A4%E4%BA%8Ep%E3%80%81q%E4%B8%A4%E7%82%B9%2C%E7%BA%BF%E6%AE%B5PQ%E7%9A%84%E4%B8%AD%E7%82%B9%E4%B8%BAM%2C%E5%8F%88l1%E4%B8%8El2%EF%BC%9Ax%2B2y%2B2%3D0%E7%9A%84%E4%BA%A4%E7%82%B9%E4%B8%BAN%2C%E6%B1%82%E8%AF%81%EF%BC%9A%7CAM%7C%2A%7CAN%7C%E4%B8%BA%E5%AE%9A%E5%80%BC)
已知圆C(x-3)²+(y-4)²=4,直线l1过定点A(1,0)若l1与圆相交于p、q两点,线段PQ的中点为M,又l1与l2:x+2y+2=0的交点为N,求证:|AM|*|AN|为定值
已知圆C(x-3)²+(y-4)²=4,直线l1过定点A(1,0)
若l1与圆相交于p、q两点,线段PQ的中点为M,又l1与l2:x+2y+2=0的交点为N,求证:|AM|*|AN|为定值
已知圆C(x-3)²+(y-4)²=4,直线l1过定点A(1,0)若l1与圆相交于p、q两点,线段PQ的中点为M,又l1与l2:x+2y+2=0的交点为N,求证:|AM|*|AN|为定值
设l1:x=ky+1,N点坐标(XN,YN),联立l1,l2得XN=-3·k/(k+2)+1,YN=-3/(k+2),则丨AN丨=3·(k2⃣️+1)1/2/(k+2)
设M(Xo,yo)P(Xl,Yl)Q(X2,y2)联立圆与丨1得(k2⃣️+1)y2⃣️-(8+4k)y+16=0
所以y1+y2=(8+4k)/(K2⃣️+1)
所以y.=(4+2K)/(K2⃣️+1) x.=1+(4+2k)K/(K2⃣️+1)
所以丨AM丨=(4+2K)·(K2⃣️+1)1/2/(K2⃣️+1)
所以丨AM丨·丨AN丨=6为定值
x1+x2=[2k(k+4)+6]/(1+k^2), x1x2=[(k+4)^2+5]/(1+k^2). ---①
联立L1与L2得 N(2(k-1)/(2k+1),-3k/(2k+1)),
验算(AM*AN)^2与k无关,
AM^2=[(x1+x2)/2-1]^2+[k(x1+x2)/2-k]^2,
AN^2=[2(k-1)/(2k+1)-1]^2+[-3k/(2k+1)]^2, ----②
由①②可 得,(AM*AN)^2=9.所以为定值