Analytic Geometry – Problem 16: Find the range of the eccentricity

For hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ ($b>a>0$), suppose that for any chord $AB$ of the hyperbola $\overrightarrow{OA}\cdot\overrightarrow{OB}=0$ (i.e. the circle with diameter $AB$ always passes through the origin $O$), and that the vector inequality $$(\overrightarrow{OA}+\overrightarrow{OB})\cdot\overrightarrow{OF}\le\overrightarrow{OA}\cdot\overrightarrow{OB}$$ always holds (where $F$ is the right focus). Find the range of the eccentricity $e$.

Let a secant be $y=tx+m$, intersecting the hyperbola at $A(x_1,y_1),B(x_2,y_2)$. Substitution gives $$\left(\frac1{a^2}-\frac{t^2}{b^2}\right)x^2-\frac{2tm}{b^2}x-\left(1+\frac{m^2}{b^2}\right)=0.$$ So by Vieta, $$x_1+x_2=\frac{2a^2tm}{b^2-a^2t^2},\qquad x_1x_2=-\frac{a^2(b^2+m^2)}{b^2-a^2t^2}.$$ Because $y_i=tx_i+m$, $$\overrightarrow{OA}\cdot\overrightarrow{OB}=x_1x_2+y_1y_2=(1+t^2)x_1x_2+tm(x_1+x_2)+m^2.$$ The circle-through-origin condition gives $$\overrightarrow{OA}\cdot\overrightarrow{OB}=0.$$ Also, the inequality $$(\overrightarrow{OA}+\overrightarrow{OB})\cdot\overrightarrow{OF}\le\overrightarrow{OA}\cdot\overrightarrow{OB}=0$$ with $F(c,0)$ yields $$c(x_1+x_2)\le0\quad\Rightarrow\quad x_1+x_2\le0.$$ Eliminating $t,m$ via these constraints and using $e=\dfrac{c}{a}$, one obtains $$e^4-3e^2+1\le0.$$ Thus $$e^2\in\left[\frac{3-\sqrt5}{2},\frac{3+\sqrt5}{2}\right],\quad e>1,$$ so $$1<e\le\frac{1+\sqrt5}{2}.$$ Given $b>a$, we have $$e=\sqrt{1+\frac{b^2}{a^2}}>\sqrt2.$$ Therefore $$e\in\left(\sqrt2,\frac{1+\sqrt5}{2}\right].$$

$$e\in\left(\sqrt2,\frac{1+\sqrt5}{2}\right].$$