题目

（20分）已知 $P(x_{0},y_{0})$ 为椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1(a>b>0)$ 上一点．

【难度】

【出处】

2013年湖南省高中数学竞赛

【标注】

设直线 $l$ 为过点 $P$ 的椭圆的切线，试求过椭圆焦点 $F(-c,0)$ 且垂直于 $l$ 的直线方程；
标注
答案

$\dfrac{y_{0}x}{b^{2}}-\dfrac{x_{0}y}{a^{2}}-\dfrac{y_{0}c}{b^{2}}=0$

解析

易知直线 $l$ 的方程为\[\dfrac{x_{0}x}{a^{2}}+\dfrac{y_{0}y}{b^{2}}=1\cdots\cdots\text{ ① }\]设过焦点 $F(-c,0)$ 且垂直于 $l$ 的直线方程为 $\dfrac{y_{0}y}{b^{2}}-\dfrac{x_{0}y}{a^{2}}+F=0$，将 $F(-c,0)$ 代入方程得 $F=-\dfrac{y_{0}c}{b^{2}}$，故要求的方程为\[\dfrac{y_{0}x}{b^{2}}-\dfrac{x_{0}y}{a^{2}}-\dfrac{y_{0}c}{b^{2}}=0\cdots\cdots\text{ ② }\]
求证：椭圆的焦点在椭圆切线上的射影的轨迹是以椭圆中心为圆心，且过长轴顶点的圆．
标注
答案

略

解析

先考察左焦点 $F(-c,0)$ 在切线上的射影．由 ①② 联立解得\[\begin{cases}x=\dfrac{1}{D}\left(\dfrac{x_{0}}{a^{2}}+\dfrac{y_{0}^{2}c}{b^{4}}\right).\\ y=\dfrac{1}{D}\left(\dfrac{y_{0}}{a^{2}}+\dfrac{x_{0}y_{0}c}{a^{2}b^{2}}\right).\end{cases}\]其中 $D=\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{y_{0}^{2}}{b^{4}}$．\[\begin{split}x^{2}+y^{2}&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}}{a^{2}}+\dfrac{y_{0}^{2}c}{b^{4}}\right)^{2}+\left(\dfrac{y_{0}}{b^{2}}-\dfrac{x_{0}y_{0}c}{a^{2}b^{2}}\right)^{2}\right]\\&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{2x_{0}y_{0}^{2}c}{a^{2}b^{4}}+\dfrac{y_{0}^{4}c^{2}}{b^{8}}\right)+\left(\dfrac{y_{0}^{2}}{b^{4}}-\dfrac{2x_{0}y_{0}^{2}c}{a^{2}b^{4}}+\dfrac{x_{0}^{2}y_{0}^{2}c^{2}}{a^{4}b^{4}}\right)\right]\\&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{y_{0}^{2}}{b^{4}}\right)+\dfrac{y_{0}^{2}c^{2}}{b^{4}}\left(\dfrac{y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{4}}\right)\right]\\&=\dfrac{1}{D^{2}}\left[\dfrac{y_{0}^{2}}{b^{4}}+1\right]\\&=\dfrac{1}{D}\left[\dfrac{y_{0}^{2}}{b^{4}}(a^{2}-b^{2})+1\right]\\&=\dfrac{1}{D}\left[\dfrac{a^{2}y_{0}^{2}}{b^{4}}+\left(1-\dfrac{y_{0}^{2}}{b^{2}}\right)\right]\cdots\cdots\text{ ③ }\end{split}\]因为 $P(x_{0},y_{0})$ 为椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}(a>b>0)$ 上一点，所以 $\dfrac{x_{0}^{2}}{a^{2}}+\dfrac{y_{0}^{2}}{b^{2}}=1$，即\[1-\dfrac{y_{0}^{2}}{b^{2}}=\dfrac{x_{0}^{2}}{a^{2}}\cdots\cdots\text{ ④ }\]将 ④ 代入 ③，得\[x^{2}+y^{2}=\dfrac{1}{D}\left[\dfrac{a^{2}y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{2}}\right]=\dfrac{a^{2}}{D}\left(\dfrac{y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{4}}\right)=a^{2}.\]对于右焦点 $F(c,0)$ 在切线上的射影，同理可证．

题目问题1 答案1 解析1 备注1 问题2 答案2 解析2

先考察左焦点 $F(-c,0)$ 在切线上的射影．由 ①② 联立解得\[\begin{cases}x=\dfrac{1}{D}\left(\dfrac{x_{0}}{a^{2}}+\dfrac{y_{0}^{2}c}{b^{4}}\right).\\ y=\dfrac{1}{D}\left(\dfrac{y_{0}}{a^{2}}+\dfrac{x_{0}y_{0}c}{a^{2}b^{2}}\right).\end{cases}\]其中 $D=\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{y_{0}^{2}}{b^{4}}$．\[\begin{split}x^{2}+y^{2}&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}}{a^{2}}+\dfrac{y_{0}^{2}c}{b^{4}}\right)^{2}+\left(\dfrac{y_{0}}{b^{2}}-\dfrac{x_{0}y_{0}c}{a^{2}b^{2}}\right)^{2}\right]\\&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{2x_{0}y_{0}^{2}c}{a^{2}b^{4}}+\dfrac{y_{0}^{4}c^{2}}{b^{8}}\right)+\left(\dfrac{y_{0}^{2}}{b^{4}}-\dfrac{2x_{0}y_{0}^{2}c}{a^{2}b^{4}}+\dfrac{x_{0}^{2}y_{0}^{2}c^{2}}{a^{4}b^{4}}\right)\right]\\&=\dfrac{1}{D^{2}}\left[\left(\dfrac{x_{0}^{2}}{a^{4}}+\dfrac{y_{0}^{2}}{b^{4}}\right)+\dfrac{y_{0}^{2}c^{2}}{b^{4}}\left(\dfrac{y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{4}}\right)\right]\\&=\dfrac{1}{D^{2}}\left[\dfrac{y_{0}^{2}}{b^{4}}+1\right]\\&=\dfrac{1}{D}\left[\dfrac{y_{0}^{2}}{b^{4}}(a^{2}-b^{2})+1\right]\\&=\dfrac{1}{D}\left[\dfrac{a^{2}y_{0}^{2}}{b^{4}}+\left(1-\dfrac{y_{0}^{2}}{b^{2}}\right)\right]\cdots\cdots\text{ ③ }\end{split}\]因为 $P(x_{0},y_{0})$ 为椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}(a>b>0)$ 上一点，所以 $\dfrac{x_{0}^{2}}{a^{2}}+\dfrac{y_{0}^{2}}{b^{2}}=1$，即\[1-\dfrac{y_{0}^{2}}{b^{2}}=\dfrac{x_{0}^{2}}{a^{2}}\cdots\cdots\text{ ④ }\]将 ④ 代入 ③，得\[x^{2}+y^{2}=\dfrac{1}{D}\left[\dfrac{a^{2}y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{2}}\right]=\dfrac{a^{2}}{D}\left(\dfrac{y_{0}^{2}}{b^{4}}+\dfrac{x_{0}^{2}}{a^{4}}\right)=a^{2}.\]对于右焦点 $F(c,0)$ 在切线上的射影，同理可证．