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[TEX]+) M \in (C) \Rightarrow M(m;\frac{2m-1}{m+1})[/TEX]
[TEX]y'=\frac{3}{(x+1)^2} \Rightarrow y'(m)=\frac{3}{(m+1)^2}[/TEX]
PTTT (d) tại M của (C): [TEX]y=\frac{3}{(m+1)^2}(x-m)+\frac{2m-1}{m+1} \Leftrightarrow 3x-(m+1)^2y+2m^2-2m+1=0[/TEX]
[TEX]d(I,(d))=\frac{|-3-2(m+1)^2+2m^2-2m+1|}{\sqrt{9+(m+1)^4}}=\frac{4|m+1|}{\sqrt{9+(m+1)^2}}=\frac{4}{\sqrt{\frac{9}{(m+1)^2}+(m+1)^2}}[/TEX]
[TEX]\frac{9}{(m+1)^2}+(m+1)^2 \geq 2.\sqrt{\frac{9}{(m+1)^2}.(m+1)^2}=6[/TEX]
[TEX]d(I,(d)) \leq \frac{4}{\sqrt{6}} \Rightarrow Maxd(I,(d))=\frac{4}{\sqrt{6}} \Leftrightarrow \frac{9}{(m+1)^2}=(m+1)^2}[/TEX]
 
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