论坛 › 趣题妙解 › 这样的直线是否一定是切线呢？

manthanein 发表于 2016-12-1 23:49:30

\(\abs{\D \frac{\psi(u)-\psi(a)}{\phi(u)-\phi(a)}-C} \lt \lambda_1\)
\(-\lambda_1 \lt \D \frac{\psi(u)-\psi(a)}{\phi(u)-\phi(a)}-C \lt \lambda_1\)
\(C-\lambda_1 \lt \D \frac{\psi(u)-\psi(a)}{\phi(u)-\phi(a)} \lt C+\lambda_1\)

manthanein 发表于 2016-12-2 21:20:23

记\(\beta=\D \frac{\psi(u)-\psi(a)}{\phi(u)-\phi(a)}\)

manthanein 发表于 2016-12-2 21:31:59

令\(X=\phi(u)-\phi(a)\)，\(Y=\psi(u)-\psi(a)\)
\(\beta=\D \frac{Y}{X}\)
\(F(u)=\sqrt{X^2+Y^2}\)
\(G(u)=\D \frac{\D \abs{AX+BY}}{\sqrt{A^2+B^2}\cdot \sqrt{X^2+Y^2}}\)

manthanein 发表于 2016-12-2 21:44:12

\(F(u)G(u)=\D \frac{\D \abs{AX+BY}}{\sqrt{A^2+B^2}}\)
\(F^2(u)G^2(u)=\D \frac{\D (AX+BY)^2}{A^2+B^2}\)
\(F^2(u)-F^2(u)G^2(u)=(X^2+Y^2)-\D \frac{\D (AX+BY)^2}{A^2+B^2}\)
\(F^2(u)-F^2(u)G^2(u)=\D \frac{\D (X^2+Y^2)(A^2+B^2)-(AX+BY)^2}{A^2+B^2}\)
\(F^2(u)-F^2(u)G^2(u)=\D \frac{\D (AY-BX)^2}{A^2+B^2}\)

manthanein 发表于 2016-12-2 21:48:15

\(\beta-k\)
\(=\D \frac{Y}{X}+\frac{A}{B}\)
\(=\D \frac{AX+BY}{BX}\)

manthanein 发表于 2016-12-2 21:51:04

\(1+\beta k\)
\(=1-\D \frac{Y}{X} \cdot \frac{A}{B}\)
\(=\D \frac{BX-AY}{BX}\)

manthanein 发表于 2016-12-2 22:11:21

\(\D \frac{F^2(u)-F^2(u)G^2(u)}{F^2(u)G^2(u)}=\frac{(BX-AY)^2}{(AX+BY)^2}\)
\(\D \frac{1+\beta k}{\beta-k}=\frac{BX-AY}{AX+BY}\)
所以：
\(\D \frac{1}{G^2(u)}-1=\frac{(1+\beta k)^2}{(\beta-k)^2}=\frac{1+\beta^2k^2+2\beta k}{\beta^2+k^2-2\beta k}\)

manthanein 发表于 2016-12-2 22:21:48

\(\D \frac{1}{G^2(u)}\)
\(=1+\D \frac{1+\beta^2k^2+2\beta k}{\beta^2+k^2-2\beta k}\)
\(=\D \frac{1+\beta^2k^2+\beta^2+k^2}{\beta^2+k^2-2\beta k}\)
\(=\D \frac{(\beta^2+1)(k^2+1)}{(\beta-k)^2} \)

\(G^2(u)=\D \frac{(\beta-k)^2}{(\beta^2+1)(k^2+1)}\)
\(G(u)=\D \frac{\D \abs{\beta-k}}{\sqrt{(\beta^2+1)(k^2+1)}}\)

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