有关圆的问题。现在就要
已知动圆过定点(p/2,0),且与直线x=-p/2相切,其中p>0. 1.求动圆心C的轨迹方程: 设A.B是轨迹C上异于原点O的两个不同点,直线OA和OB的倾斜角分别为A,B,当A ,B变化且A+B=45度时,证明直线AB恒过定点。并求出该定点的坐标
1. 设圆心坐标C(x,y), 点C到顶点的距离 = 到直线的距离: 根号[(x -p/2)^2+y^2] = |x +p/2| ===> y^2 = 2px 此即为圆心C的轨迹方程 2. 设:A(a^2/2p,a), B(b^2/2p,b) tanA = a/(a^2/2p) =2p/a, tanB = 2p/b tan(A+B)=tan45 =1 = (tanA+tanB)/(1-tanAtanB) ==> ab = 2p(a+b)+4*p^2 ...(1) 直线AB: (y -a)/(x -a^2/2p) = (b-a)/(b^2/2p -a^2/2p) ===> y-2p = 2p(x+2p)/(a+b) 因此,直线AB恒过定点(-2p,2p)
. 设圆心坐标C(x,y), 点C到顶点的距离 = 到直线的距离: 根号[(x -p/2)^2+y^2] = |x +p/2| ===> y^2 = 2px 此即为圆心C的轨迹方程 2. 设:A(a^2/2p,a), B(b^2/2p,b) tanA = a/(a^2/2p) =2p/a, tanB = 2p/b tan(A+B)=tan45 =1 = (tanA+tanB)/(1-tanAtanB) ==> ab = 2p(a+b)+4*p^2 ...(1) 直线AB: (y -a)/(x -a^2/2p) = (b-a)/(b^2/2p -a^2/2p) ===> y-2p = 2p(x+2p)/(a+b) 因此,直线AB恒过定点(-2p,2p
解:∵2a1、S(n+1)、Sn成等差数列 ∴2a1+Sn=2S(n+1) 即Sn=2S(n+1)-2 ∴S(n-1)=2Sn-2 ∴Sn-S(n-1)=2S(n+1)-2-2Sn+2 =2[S(n+1)-Sn] ∴S(n+1)-Sn/Sn-S(n-1)=1/2 ∴{Sn-S(n-1)}是以a1=1为首项,1/2为公比的等比数列 则有an=Sn-S(n-1)=a1×q^n-1=(1/2)^(n-1)
答:解:设动圆C的圆心C坐标为(x,y) ∵动圆过定点F(1/2,0)且与定直线l:x=-1/2相切 ∴C点到L距离d=x+1/2 C点到F距离d1=√[(x-1/...详情>>
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