1009

已知椭圆 ${\displaystyle M:{\displaystyle \frac{M^2}{a^2}+y^2=1}}$, 圆 ${\displaystyle C:x^2+y^2=6-a^2}$ 在第一象限有公共点 ${\displaystyle P}$, 设圆 ${\displaystyle C}$ 在点 ${\displaystyle P}$ 处的切线斜率为 ${\displaystyle k}$, 椭圆 ${\displaystyle M}$ 在点 ${\displaystyle P}$ 处的切线斜率为 ${\displaystyle k_2}$, 求 ${\displaystyle {\displaystyle \frac{k_1}{k_2}}}$ 的取值范围.

联立 ${\displaystyle M,C}$,得交点坐标 ${\displaystyle (x_0,y_0)}$ 满足$$\begin{array}{r}{x}_0^2=\frac{a^2(5-a^2)}{a^2-1},y_0^2=\frac{2(a^2-3)}{a^2-1}>0,\end{array}$$
解得${\displaystyle 3<a^2<5}$. 进而${\displaystyle k_1=-{\displaystyle \frac{x_0}{y_0},k_2=-\frac{x_0}{a^2{y}_0}}}$, 故$$\begin{array}{r}\frac{k_1}{k_2}=a^2\in (3,5).\end{array}$$