(1-tanp)*x=1 x=(1-tanp)/2 所以M((1-tanp)/2 |BN|^2=2[tanp/2 y=(1+tanp)/,tanp/:y=tanp*x -x+1=tanp*x x=1/,1) ∠BON=p 直线AB;(1-tanp)*x 2/:y=-x+1 直线NC;2) |AM|^2=(1-tanp)^2/2-tanp/(1+tanp);2(1+tanp)^2-(1-tanp)^2/,0) B(1;(1+tanp) 所以N(1/令C(0;2-1/4)*x=(tanp+1)/(1+tanp) y=tanp/(1+tanp)) 直线MC;(1+tanp)]^2 所以|MN|^2-|AM|^2-|BN|^2 =[1+(tanp)^2]^2/2-2(tanp)^2/:y=tan(p+π/2;(1-tanp)*x -x+1=(tanp+1)/(1+tanp)]^2 |MN|^2=[(1-tanp)/,(1+tanp)/(1+tanp)]^2+[(1+tanp)/(1+tanp)^2 =[1+2(tanp)^2+(tanp)^4-1+2(tanp)^2-(tanp)^4-4(tanp)^2]/,0) A(0 crs0723 2012-5-7 |
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