46409 關於雙曲線與橢圓的兩個命題

\begin{align} S_{\triangle ACB}=\frac 34 S_{弓}=\frac 12 S_{\triangle APB},\label{1} \end{align}

\begin{align} S_{\triangle ACB}\gt\frac 34 S_{弓}\gt\frac 12 S_{\triangle APB}.\label{2} \end{align}

\begin{align} S_{\triangle ACB}=\frac 12\left| \begin{array}{ccccc} ~a~&&\dfrac 1a&&~1~\\[6pt] \sqrt{ab}&&\dfrac 1{\sqrt{ab}}&&~1~\\[6pt] ~b~&&\dfrac 1b&&~1~ \end{array} \right|=\frac{b-a}{2ab}\Big(\sqrt b-\sqrt a\Big)^2.\label{3} \end{align}

\begin{align*} y-\frac 1a&=-\frac 1{a^2}(x-a),\\ y-\frac 1b&=-\frac 1{b^2}(x-b). \end{align*}

\begin{align*} S_{\triangle APB}=\frac 12\left| \begin{array}{ccccc} ~a~&&\dfrac 1a&&~1~\\[6pt] ~\dfrac{2ab}{a+b}~&&\dfrac 2{a+b}&&~1~\\[6pt] ~b~&&\dfrac 1b&&~1~ \end{array}\right|=\frac{1}{2(a+b)} \left| \begin{array}{ccccc} ~a~&&\dfrac 1a&&~1~\\[6pt] ~2ab~&&2&&~a+b~\\[6pt] ~b~&&\dfrac 1b&&~1~ \end{array}\right|, \end{align*}

\begin{align} S_{\triangle APB}&=\dfrac{(b-a)^3}{2ab(a+b)}.\label{5} \end{align}

\begin{align} &\frac{b-a}{2ab}(\sqrt b-\sqrt a)^2\gt\frac 34\Big(\frac{b^2-a^2}{2ab}-\ln \frac ba\Big),\nonumber\\ {\hbox{或}} &\frac{b-a}{ab}(\sqrt b-\sqrt a)^2-\frac 34\frac{b^2-a^2}{ab}+\frac 32 \ln \frac ba\gt0,\nonumber\\ {\hbox{或}} &\dfrac{\frac ba-1}{\frac ba}\Big(\sqrt{\frac ba}-1\Big)^2-\frac 34\dfrac{(\frac ba)^2-1}{\frac ba}+\frac 32 \ln \frac ba\gt0.\label{6} \end{align}

\begin{align} &\frac{x-1}{x}(\sqrt x-1)^2-\frac 34\frac{x^2-1}{x}+\frac 32 \ln x\gt0,\label{7}\\ {\hbox{或}} &x-\frac 1x-2\sqrt{x}+\frac 2{\sqrt{x}}-\frac 34 x+\frac 34\frac 1x+\frac 32\ln x\gt0,\nonumber\\ {\hbox{即}} &x-\frac 1x-8\sqrt{x}+\frac 8{\sqrt{x}}+6\ln x\gt0.\label{8} \end{align}

$$g'(x)=1+\frac 1{x^2}-\frac 4{\sqrt x}-\frac 4{x\sqrt x}+\frac 6x=\frac 1{x^2}(\sqrt{x}-1)^4.$$

\begin{align} S_{\triangle ACB}=\frac 34S_{弓}.\label{9} \end{align}

\begin{align} &\frac 34\Big(\frac{b^2-a^2}{2ab}-\ln \frac ba\Big)\gt\frac 12\cdot \frac{(b-a)^3}{2ab(a+b)},\nonumber\\ {\hbox{或}} &\frac{b^2-a^2}{2ab}-\frac{(b-a)^3}{3ab(a+b)}-\ln \frac ba\gt0.\label{10} \end{align}

\begin{align} &\frac{x^2-1}{2x}-\frac{(x-1)^3}{3x(x+1)}-\ln x\gt0,\label{11}\\ {\hbox{或}} &\frac 1{6x(x+1)}(x^3+9x^2-9x-1)-\ln x\gt0.\label{12} \end{align}

$$h'(x)=\frac{1}{6x^2(x+1)^2}(x^4-4x^3+6x^2-4x+1)=\frac{(x-1)^4}{6x^2(x+1)^2},$$

\begin{align} \frac 34 S_{弓}\gt\frac 12 S_{\triangle APB}.\label{13} \end{align}

\begin{align} S_{\triangle ACB}\lt \frac 34S_{弓}\lt\frac 12 S_{\triangle APB}.\label{14} \end{align}

\begin{align*} S_{\triangle ABT}&=\frac 12\sin\theta_1+\frac 12\sin\theta_2-S_{\triangle AOB}\\ &=\sin\theta\cos\frac 12(\theta_1-\theta_2)-\sin\theta\cos\theta, \end{align*}

\begin{align} S_{\triangle ACB}=\sin\theta-\sin\theta\cos\theta=\sin\theta(1-\cos\theta).\label{15} \end{align}

\begin{align} S_{弓}&=\theta-\sin\theta\cos\theta,\label{16}\\ S_{\triangle APB}&=\frac 12\cdot 2\sin\theta\cdot \frac{\sin^2\theta}{\cos\theta}= \frac{\sin^3\theta}{\cos\theta}.\label{17} \end{align}

\begin{align} &\sin\theta-\sin\theta\cos\theta\lt\frac 34\theta-\frac 34\sin\theta\cos\theta,\nonumber\\ {\hbox{亦即}} &\frac 34\theta+\dfrac 14\sin\theta\cos\theta-\sin\theta\gt0.\label{18} \end{align}

$$u'(\theta)=\frac 34+\frac 14\cos 2\theta-\cos\theta=\frac 12(1-\cos\theta)^2,$$

\begin{align} S_{\triangle ACB}\lt \frac 34S_{弓}.\label{19} \end{align}

\begin{align} \theta-\sin\theta\cos\theta-\dfrac 23\frac{\sin^3\theta}{\cos\theta}\lt0. \label{20} \end{align}

$$v'(\theta)=1-\cos 2\theta-\frac 23\cdot \frac 1{\cos^2\theta}(3\sin^2\theta\cos^2\theta+\sin^4\theta)=-\frac 23\cdot\frac{\sin^4\theta}{\cos^2\theta},$$

\begin{align} \frac 34S_{弓}\lt\frac 12 S_{\triangle APB}.\label{21} \end{align}