# 来自西哥的一道解几题

Eufisky posted @ 2014年7月27日 02:59 in 解析几何 with tags 解析几何 , 1159 阅读

(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
$\left\{ \begin{array}{l}{x_1} = 5\cos \alpha \\{y_1} = 4\sin \alpha\end{array} \right.\text{及}\left\{ \begin{array}{l}{x_2} = 5\cos \beta \\{y_2} = 4\sin \beta\end{array} \right.$

$\frac{{4\sin \beta - 4\sin \alpha }}{{5\cos \beta - 5\cos \alpha }} = \frac{{4\sin \beta }}{{5\cos \beta + 3}} \Leftrightarrow 3\left( {\sin \beta - \sin \alpha } \right) = 5\sin \left( {\alpha - \beta } \right).$
$5\cos \frac{{\alpha - \beta }}{2} = - 3\cos \frac{{\alpha + \beta }}{2}\Leftrightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2} = - 4.$

\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*}

$\left\{ \begin{array}{l}x = \frac{{\left| {B{F_2}} \right|{x_1} + \left| {A{F_2}} \right|{x_2} + 3\left| {AB} \right|}}{{20}}\\y = \frac{{\left| {B{F_2}} \right|{y_1} + \left| {A{F_2}} \right|{y_2}}}{{20}}\end{array} \right.$

$\left\{ \begin{array}{l}x = \frac{{34\left( {\cos \beta + \cos \alpha } \right) - 30\cos \alpha \cos \beta + 30}}{{20}}\\y = \frac{{5\left( {\sin \beta + \sin \alpha } \right) - 3\sin \left( {\alpha + \beta } \right)}}{5}\end{array} \right.$

\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*}

$\left\{ \begin{array}{l}x = - \frac{{51}}{{25}}\cos \left( {\alpha + \beta } \right) + \frac{{24}}{{25}}\\y = - \frac{6}{5}\sin \left( {\alpha + \beta } \right)\end{array} \right. \Rightarrow {\left( {\frac{{25x - 24}}{{51}}} \right)^2} + \frac{{25}}{{36}}{y^2} = 1\left( {x = \frac{{27}}{{25}},y = 0} \right).$
(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*}

$r = \frac{4}{{\left| {t + \frac{1}{t}} \right|}} = \frac{4}{{\left| t \right| + \frac{1}{{\left| t \right|}}}} \le \frac{4}{{\frac{4}{3} + \frac{3}{4}}} = \frac{{48}}{{25}}.$

$\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = - \frac{{16}}{5}\end{array} \right.\text{或}\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = \frac{{16}}{5}\end{array} \right.$

则$$x_{1}+x_{2}=-\dfrac{2a^2ck^2}{b^2+a^2k^2},x_{1}x_{2}=\dfrac{a^2c^2k^2-a^2b^2}{b^2+a^2k^2}$$

因为$$|AF_{1}|=a+ex_{1},|AF_{2}|=a-ex_{1},|BF_{1}|=a+ex_{2},|BF_{2}|=a-ex_{2}$$

\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*}

\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*}

$$(a^2+c^2)^2y^2=(b^2c-b^2x)(a^2x+c^3)=b^2(cb^2x+c^4-a^2x)$$
$$a^2b^2x^2+(a^2+c^2)^2y^2-cb^4x-b^2c^4=0$$

(2)设直线$AB$方程为$x=ky-3$与椭圆联立得$$(16k^2+25)y^2-96ky-256=0$$

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