已知椭圆 $E:\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1(a>b>0)$ 的一个焦点为 $F_{1}(-\sqrt 3,0)$，且过点 $H\left(\sqrt 3,\dfrac{1}{2}\right)$．设椭圆 $E$ 的上下顶点分别为 $A_1,A_2$，点 $P$ 是椭圆上异于 $A_1,A_2$ 的任一点，直线 $PA_1,PA_2$ 分别交 $x$ 轴于点 $M,N$，若直线 $OT$ 与过点 $M,N$ 的圆 $G$ 相切，切点为 $T$．证明：线段 $OT$ 的长为定值，并求出该定值．
【难度】
【出处】
2013年全国高中数学联赛甘肃省预赛
【标注】
【答案】
证明略，定值为 $2$
【解析】
由题意得 $a^{2}-b^{2}=3$，$\dfrac{3}{a^{2}}+\dfrac{1}{4b^{2}}=1$，解得 $a^{2}=4$，$b^{2}=1$，所以椭圆 $E$ 的方程为\[\dfrac{x^{2}}{4}+y^{2}=1.(*)\]解法一 由 $(*)$ 知 $A_{1}(0,1)$，$A_{2}(0,-1)$，设 $P(x_{0},y_{0})$．
直线 $PA_{1}:y-1=\dfrac{y_{0}-1}{x_{0}}x$，令 $y=0$，得 $x_{M}=\dfrac{-x_{0}}{y_{0}-1}$；
直线 $PA_{2}:y+1=\dfrac{y_{0}+1}{x_{0}}x$，令 $y=0$，得 $x_{N}=\dfrac{x_{0}}{y_{0}+1}$．
设圆 $G$ 的圆心为 $\left(\dfrac{1}{2}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right),h\right)$，则\[\begin{split}r^{2}&=\left[\dfrac{1}{2}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)-\dfrac{x_{0}}{y_{0}+1}\right]^{2}+h^{2}\\&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}+\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}.\\ OG^{2}&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}.\\ OT^{2}&=OG^{2}-r^{2}\\&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}-\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}+\dfrac{x_{0}}{y_{0}-1}\right)^{2}-h^{2}\\&=\dfrac{x_{0}^{2}}{1-y_{0}^{2}}.\end{split}\]而 $\dfrac{x_{0}^{2}}{4}+y_{0}^{2}=1$，所以 $x_{0}^{2}=4(1-y_{0}^{2})$，$OT^{2}=\dfrac{4(1-y_{0}^{2}}{1-y_{0}^{2}}=4$．所以 $|OT|=2$，即线段 $OT$ 的长度为定值 $2$．
解法二 由 $(*)$ 知 $A_{1}(0,1)$，$A_{2}(0,-1)$，设 $P(x_{0},y_{0})$．
直线 $PA_{1}:y-1=\dfrac{y_{0}-1}{x_{0}}x$，令 $y=0$，得 $x_{M}=\dfrac{-x_{0}}{y_{0}-1}$；
直线 $PA_{2}:y+1=\dfrac{y_{0}+1}{x_{0}}x$，令 $y=0$，得 $x_{N}=\dfrac{x_{0}}{y_{0}+1}$．
因此 $|OM|\cdot|ON|=\left|\dfrac{x_0}{y_0+1}\cdot\dfrac{-x_0}{y_0-1}\right|=\left|\dfrac{x_0^2}{y_0^2-1}\right|$，而 $\dfrac{x_0^2}{4}+y_0^2=1$，所以 $|OM|\cdot|ON|=4$．由切割线定理，得\[OT^2= |OM|\cdot|ON|=4,\]所以 $|OT|=2$．
直线 $PA_{1}:y-1=\dfrac{y_{0}-1}{x_{0}}x$，令 $y=0$，得 $x_{M}=\dfrac{-x_{0}}{y_{0}-1}$；
直线 $PA_{2}:y+1=\dfrac{y_{0}+1}{x_{0}}x$，令 $y=0$，得 $x_{N}=\dfrac{x_{0}}{y_{0}+1}$．
设圆 $G$ 的圆心为 $\left(\dfrac{1}{2}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right),h\right)$，则\[\begin{split}r^{2}&=\left[\dfrac{1}{2}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)-\dfrac{x_{0}}{y_{0}+1}\right]^{2}+h^{2}\\&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}+\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}.\\ OG^{2}&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}.\\ OT^{2}&=OG^{2}-r^{2}\\&=\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}-\dfrac{x_{0}}{y_{0}-1}\right)^{2}+h^{2}-\dfrac{1}{4}\left(\dfrac{x_{0}}{y_{0}+1}+\dfrac{x_{0}}{y_{0}-1}\right)^{2}-h^{2}\\&=\dfrac{x_{0}^{2}}{1-y_{0}^{2}}.\end{split}\]而 $\dfrac{x_{0}^{2}}{4}+y_{0}^{2}=1$，所以 $x_{0}^{2}=4(1-y_{0}^{2})$，$OT^{2}=\dfrac{4(1-y_{0}^{2}}{1-y_{0}^{2}}=4$．所以 $|OT|=2$，即线段 $OT$ 的长度为定值 $2$．
直线 $PA_{1}:y-1=\dfrac{y_{0}-1}{x_{0}}x$，令 $y=0$，得 $x_{M}=\dfrac{-x_{0}}{y_{0}-1}$；
直线 $PA_{2}:y+1=\dfrac{y_{0}+1}{x_{0}}x$，令 $y=0$，得 $x_{N}=\dfrac{x_{0}}{y_{0}+1}$．
因此 $|OM|\cdot|ON|=\left|\dfrac{x_0}{y_0+1}\cdot\dfrac{-x_0}{y_0-1}\right|=\left|\dfrac{x_0^2}{y_0^2-1}\right|$，而 $\dfrac{x_0^2}{4}+y_0^2=1$，所以 $|OM|\cdot|ON|=4$．由切割线定理，得\[OT^2= |OM|\cdot|ON|=4,\]所以 $|OT|=2$．
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