来自西哥的一道解几题
设椭圆方程为$\frac{x^2}{25}+\frac{y^2}{16}=1$，$F_1,F_2$分别是左右焦点，设椭圆上的动直线$A,B$过左焦点$F_1$.
(1)求$\triangle ABF_2$内心的轨迹；
(2)当$\triangle ABF_2$内切圆面积最大时，求直线$AB$的方程.
(1)根据题意有$F_1(-3,0),F_2(3,0)$，不妨设$A(x_1,y_1),B(x_2,y_2)$，且
\[\left\{ \begin{array}{l}{x_1} = 5\cos \alpha \\{y_1} = 4\sin \alpha\end{array} \right.\text{及}\left\{ \begin{array}{l}{x_2} = 5\cos \beta \\{y_2} = 4\sin \beta\end{array} \right.\]
由$A,B,F_1$三点共线可知
\[\frac{{4\sin \beta - 4\sin \alpha }}{{5\cos \beta - 5\cos \alpha }} = \frac{{4\sin \beta }}{{5\cos \beta + 3}} \Leftrightarrow 3\left( {\sin \beta - \sin \alpha } \right) = 5\sin \left( {\alpha - \beta } \right).\]
即
\[5\cos \frac{{\alpha - \beta }}{2} = - 3\cos \frac{{\alpha + \beta }}{2}\Leftrightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2} = - 4.\]
由椭圆第二定义可知
\begin{align*}\left| {A{F_2}} \right| &= 5 - 3\cos \alpha \\\left| {B{F_2}} \right| &= 5 - 3\cos \beta \\\left| {AB} \right| &= \left| {A{F_1}} \right| + \left| {A{F_1}} \right| = 3\left( {\cos \beta + \cos \alpha }\right) + 10.\end{align*}
记$\triangle ABF_2$内心为$(x,y)$，则
\[\left\{ \begin{array}{l}x = \frac{{\left| {B{F_2}} \right|{x_1} + \left| {A{F_2}} \right|{x_2} + 3\left| {AB} \right|}}{{20}}\\y = \frac{{\left| {B{F_2}} \right|{y_1} + \left| {A{F_2}} \right|{y_2}}}{{20}}\end{array} \right.\]
化简整理得
\[\left\{ \begin{array}{l}x = \frac{{34\left( {\cos \beta + \cos \alpha } \right) - 30\cos \alpha \cos \beta + 30}}{{20}}\\y = \frac{{5\left( {\sin \beta + \sin \alpha } \right) - 3\sin \left( {\alpha + \beta } \right)}}{5}\end{array} \right.\]
注意到
\begin{align*}\sin \beta + \sin \alpha &= 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2} = - \frac{6}{5}\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha + \beta }}{2}\\&= - \frac{3}{5}\sin \left( {\alpha + \beta } \right)\\\cos \beta + \cos \alpha &= 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2} = - \frac{6}{5}{\cos ^2}\frac{{\alpha + \beta }}{2}\\&= - \frac{3}{5}\cos \left( {\alpha + \beta } \right) - \frac{3}{5}\\\cos \alpha \cos \beta &= \frac{1}{2}\cos \left( {\alpha + \beta } \right) + \frac{1}{2}\cos \left( {\alpha - \beta } \right)\\&= \frac{1}{2}\cos \left( {\alpha + \beta } \right) + \frac{1}{2}\left( {2{{\cos }^2}\frac{{\alpha - \beta }}{2} - 1} \right)\\&= \frac{1}{2}\cos \left( {\alpha + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}{\cos ^2}\frac{{\alpha + \beta }}{2}\\&= \frac{1}{2}\cos \left( {\alpha + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}\frac{{\cos \left( {\alpha + \beta } \right) + 1}}{2}\\&= \frac{{17}}{{25}}\cos \left( {\alpha + \beta } \right) - \frac{8}{{25}}.\end{align*}
故可以再次化简成
\[\left\{ \begin{array}{l}x = - \frac{{51}}{{25}}\cos \left( {\alpha + \beta } \right) + \frac{{24}}{{25}}\\y = - \frac{6}{5}\sin \left( {\alpha + \beta } \right)\end{array} \right. \Rightarrow {\left( {\frac{{25x - 24}}{{51}}} \right)^2} + \frac{{25}}{{36}}{y^2} = 1\left( {x = \frac{{27}}{{25}},y = 0} \right).\]
(2)由题意
\[S = \frac{1}{2} \times 20r = \frac{1}{2} \times \left| {{F_1}{F_2}} \right| \times 4\left| {\sin \beta - \sin \alpha } \right| = \frac{1}{2} \times 24\left| {\sin \beta - \sin \alpha } \right|.\]
\begin{align*}\Rightarrow r &= \frac{6}{5}\left| {\sin \beta - \sin \alpha } \right| = 2\left| {\sin \left( {\alpha - \beta } \right)} \right|\\&= 4\left| {\frac{{\sin \frac{{\alpha - \beta }}{2}\cos \frac{{\alpha - \beta }}{2}}}{{{{\sin }^2}\frac{{\alpha -\beta }}{2} + {{\cos }^2}\frac{{\alpha - \beta }}{2}}}} \right|\\&= 4\left| {\frac{{\tan \frac{{\alpha - \beta }}{2}}}{{{{\tan }^2}\frac{{\alpha - \beta }}{2} + 1}}} \right|.\end{align*}
令$t = \tan \frac{{\alpha - \beta }}{2}$，则$\left| t \right| = \left| {\tan \frac{{\alpha - \beta }}{2}} \right| = \left| {\frac{{\tan \frac{\alpha }{2} - \tan \frac{\beta }{2}}}{{1 + \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}}} \right| = \frac{{\left| {\tan \frac{\alpha }{2}} \right| + \frac{4}{{\left| {\tan \frac{\alpha }{2}} \right|}}}}{3} \ge \frac{4}{3}.$由此
\[r = \frac{4}{{\left| {t + \frac{1}{t}} \right|}} = \frac{4}{{\left| t \right| + \frac{1}{{\left| t \right|}}}} \le \frac{4}{{\frac{4}{3} + \frac{3}{4}}} = \frac{{48}}{{25}}.\]
当且仅当$\left| {\tan \frac{\alpha }{2}} \right| = \left| {\tan \frac{\beta }{2}} \right| = 2$时取等成立，这时有
\[\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = - \frac{{16}}{5}\end{array} \right.\text{或}\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = \frac{{16}}{5}\end{array} \right.\]
此时直线$AB$的方程为$x=-3.$
另附西哥的解答：
解:引理:设$\Delta ABC$的三边为$a,b,c$,且顶点$A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})$, 内心坐标$I(x,y)$,则有$$\begin{cases}x=\dfrac{ax_{1}+bx_{2}+cx_{3}}{a+b+c}\\y=\dfrac{ay_{1}+by_{2}+cy_{3}}{a+b+c}\end{cases}$$
设$A(x_{1},y_{1}),B(x_{2},y_{2}),I(x,y)$,则$AB$方程为$y=k(x+c)$与椭圆联立得$$(b^2+a^2k^2)x^2+2a^2ck^2x+a^2c^2k^2-a^2b^2=0$$
则$$x_{1}+x_{2}=-\dfrac{2a^2ck^2}{b^2+a^2k^2},x_{1}x_{2}=\dfrac{a^2c^2k^2-a^2b^2}{b^2+a^2k^2}$$
所以$$y_{1}+y_{2}=k(x_{1}+x_{2}+2c)=\dfrac{2b^2ck}{b^2+a^2k^2}$$$$x_{1}y_{2}+x_{2}y_{1}=2kx_{1}x_{2}+kc(x_{1}+x_{2})=-\dfrac{2a^2b^2k}{b^2+a^2k^2}$$
因为$$|AF_{1}|=a+ex_{1},|AF_{2}|=a-ex_{1},|BF_{1}|=a+ex_{2},|BF_{2}|=a-ex_{2}$$
所以$$|AB|=|AF_{1}|+|BF_{1}|=2a+e(x_{1}+x_{2})$$
所以\begin{align*}&x=\dfrac{(a-ex_{2})x_{1}+(a-ex_{1})x_{2}+[2a+e(x_{1}+x_{2})]c}{4a}\\&=\dfrac{(a+ec)(x_{1}+x_{2})-2ex_{1}x_{2}+2ac}{4a}\\&=\dfrac{\dfrac{a^2+c^2}{a}\left(-\dfrac{2a^2ck^2}{b^2+a^2k^2}\right)-\dfrac{2c}{a}\cdot\dfrac{a^2c^2k^2-a^2b^2}{b^2+a^2k^2}+2ac}{4a(b^2+a^2k^2)}\\&=\dfrac{4ab^2c-4ac^3k^2}{4a(b^2+a^2k^2)}\\&=\dfrac{b^2c-c^3k^2}{b^2+a^2k^2}\end{align*}
所以$$k^2=\dfrac{b^2c-b^2x}{a^2x+c^3}$$
\begin{align*}y&=\dfrac{(a-ex_{2})y_{1}+(a-ex_{1})y_{2}}{4a}=\dfrac{a(y_{1}+y_{2})-e(x_{1}y_{2}+x_{2}y_{1})}{4a}\\&=\dfrac{a\cdot\dfrac{2b^2ck}{b^2+a^2k^2}-\dfrac{c}{a}\left(-\dfrac{2a^2b^2k}{b^2+a^2k^2}\right)}{4a}\\&=\dfrac{b^2ck}{b^2+a^2k^2}\end{align*}
所以
\begin{align*}y^2&=\dfrac{b^4c^2k^2}{(b^2+a^2k^2)^2}=\dfrac{b^2c^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}}{\left(b^2+a^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}\right)^2}=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{(a^2x+c^3)b^2+a^2(b^2c-b^2x)}{a^2x+c^3}\right]^2}\\&=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{b^2c(a^2+c^2)}{a^2x+c^3}\right]^2}=\dfrac{b^2(c-x)(a^2x+c^3)}{(a^2+c^2)^2}\end{align*}
所以
$$(a^2+c^2)^2y^2=(b^2c-b^2x)(a^2x+c^3)=b^2(cb^2x+c^4-a^2x)$$
即
$$a^2b^2x^2+(a^2+c^2)^2y^2-cb^4x-b^2c^4=0$$
此问题中,令$a=5,b=4$,则内心I的轨迹方程为$$100x^2+289y^2-192x=324$$
(2)设直线$AB$方程为$x=ky-3$与椭圆联立得$$(16k^2+25)y^2-96ky-256=0$$
所以$$y_{1}+y_{2}=\dfrac{96k}{16k^2+25},y_{1}y_{2}=-\dfrac{256}{16k^2+25}$$
所以$$S_{\Delta ABC}=\dfrac{1}{2}|F_{1}F_{2}||y_{1}-y_{2}|=3\sqrt{(y_{1}+y_{2})^2-4y_{1}y_{2}}=3\sqrt{\dfrac{96^2k^2}{(16k^2+25)^2}+\dfrac{4\times}{16k^2+25}}$$$$S_{\Delta ABC}=960\sqrt{\dfrac{k^2+1}{(16k^2+25)^2}}$$
令$k^2+1=t\ge1$,则有$$\dfrac{k^2+1}{(16k^2+25)^2}=\dfrac{t}{(16t+9)^2}=\dfrac{t}{256t^2+288t+81}=\dfrac{1}{256t+\dfrac{81}{t}+288}\le\dfrac{1}{625}=\dfrac{1}{25^2}$$
即$S_{\Delta ABC}\le 38.4$当且仅当$k=0$时取等.故此时$AB$方程为$x=-3$.
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