25~26高三下·陕西榆林模拟·第7题

B

易知直线$A_1A_2$的倾斜角为$\dfrac{3\pi}{8}$，

则直线$A_1A_2$的斜率$k = \tan\dfrac{3\pi}{8} = \sqrt{2} + 1$，

故直线$A_1A_2$的方程为$y = (\sqrt{2} + 1)x$.

联立
$
\begin{cases}
y = (\sqrt{2} + 1)x \ y = x + \dfrac{1}{x}
\end{cases}
$

解得$x^2 = \dfrac{\sqrt{2}}{2}$，$y^2 = \dfrac{3\sqrt{2}}{2} + 2$，

则$a^2 = |OA_2|^2 = x^2 + y^2 = 2 + 2\sqrt{2}$.

由题意可知，函数$f(x)$图象的另一条渐近线为直线$x = 0$.

过点$A_1$和点$A_2$分别作直线$A_1A_2$的垂线，分别交直线$y = x$、$x = 0$于点$N,M,Q,P$.

$
b = |A_2P| = a\tan\dfrac{\pi}{8} = a \cdot \dfrac{\sin\dfrac{\pi}{8}}{\cos\dfrac{\pi}{8}}$

$= a \cdot \dfrac{2\sin^2\dfrac{\pi}{8}}{2\sin\dfrac{\pi}{8}\cos\dfrac{\pi}{8}} = a \cdot \dfrac{1 – \cos\dfrac{\pi}{4}}{\sin\dfrac{\pi}{4}}$

$= a \cdot \dfrac{1 – \dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}} = (\sqrt{2} – 1)a
$

因此
$
ab = (\sqrt{2} – 1)a^2 $$= (\sqrt{2} – 1)(2 + 2\sqrt{2}) = 2
$

故选：B.