过点N(-3,4)作圆:(x+1)^2+(y+1)^2=4的弦交A,B两点,求A,B中点的轨迹方程求求求求求求过程.
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过点N(-3,4)作圆:(x+1)^2+(y+1)^2=4的弦交A,B两点,求A,B中点的轨迹方程求求求求求求过程.
过点N(-3,4)作圆:(x+1)^2+(y+1)^2=4的弦交A,B两点,求A,B中点的轨迹方程
求求求求求求过程.
过点N(-3,4)作圆:(x+1)^2+(y+1)^2=4的弦交A,B两点,求A,B中点的轨迹方程求求求求求求过程.
圆C:(x+1)^2+(y+1)^2=4
N(-3,4),C(-1,-1)
A,B中点P(x,y),AB⊥CP
k(CP)=(y+1)/(x+1),k(AB)=k(PN)=(y-4)/(x+3)
k(CP)*k(AB)=-1
[(y+1)/(x+1)]*[(y-4)/(x+3)]=-1
(x+2)^2+(y-1.5)^2=7.25
设y=kx+b,N(-3,4)代入得:y=kx+3k+4;与圆组合得:(k^2+1)x^2+(6k^2+10k+2)x+39k-3=0……(1)
设中点为m(x,y);(1)的求跟公式:x1,2=(-(6k^2+10k+2)+-sqrt((6k^2+10k+2)^2-4(k^2+1)(39k-3))/(2(k^2+1)));
x=x1+x2=-(6k^2+10K+2)/(K^2+1...
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设y=kx+b,N(-3,4)代入得:y=kx+3k+4;与圆组合得:(k^2+1)x^2+(6k^2+10k+2)x+39k-3=0……(1)
设中点为m(x,y);(1)的求跟公式:x1,2=(-(6k^2+10k+2)+-sqrt((6k^2+10k+2)^2-4(k^2+1)(39k-3))/(2(k^2+1)));
x=x1+x2=-(6k^2+10K+2)/(K^2+1)代入y=kx+3k+4=-k(6k^2+10K+2)/(k^2+1)+3k+4;
将(x,y)代入y=kx+3k+4
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