25~26高三下·湖北武汉三调·第8题

D

设$A(-a,0)$，$B(a,0)$，任取点列中的一点$P_n(x,y)$，

则$\dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1$，$y \neq 0$，

$\overrightarrow{P_nA} = (-a-x,-y)$，$\overrightarrow{P_nB} = (a-x,-y)$.

$
\begin{align} \overrightarrow{P_nA} \cdot \overrightarrow{P_nB} &= (-a-x)(a-x) + y^2 = x^2 + y^2 – a^2, \ \dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} &= 1 \implies x^2 – a^2 = \dfrac{a^2}{b^2}y^2, \ \therefore \overrightarrow{P_nA} \cdot \overrightarrow{P_nB} &= \dfrac{a^2 + b^2}{b^2}y^2. \end{align}
$

在$\triangle AP_nB$中，$AB = 2a$，高为$|y|$，

$
\dfrac{1}{2}|\overrightarrow{P_nA}| \cdot |\overrightarrow{P_nB}| \sin\theta_n = a|y|,
$

$
\therefore |\overrightarrow{P_nA}| \cdot |\overrightarrow{P_nB}| \sin\theta_n = 2a|y|,
$

$
\therefore |\overrightarrow{P_nA}|^2 |\overrightarrow{P_nB}|^2 = (\overrightarrow{P_nA} \cdot \overrightarrow{P_nB})^2 + 4a^2y^2.
$

$
\begin{align} \dfrac{1}{1-\cos2\theta_n} &= \dfrac{1}{2\sin^2\theta_n} = \dfrac{|\overrightarrow{P_nA}|^2 |\overrightarrow{P_nB}|^2}{8a^2y^2} \ &= \dfrac{1}{2} + \dfrac{(\overrightarrow{P_nA} \cdot \overrightarrow{P_nB})^2}{8a^2y^2} \ &= \dfrac{1}{2} + \dfrac{a^2 + b^2}{8a^2b^2} \left( \overrightarrow{P_nA} \cdot \overrightarrow{P_nB} \right). \end{align}
$

设${\overrightarrow{P_nA} \cdot \overrightarrow{P_nB}}$的公差为$d$，则$\left{\dfrac{1}{1-\cos2\theta_n}\right}$的公差为$\dfrac{a^2 + b^2}{8a^2b^2}d$.

又两数列列公差相等，
$
\therefore \dfrac{a^2 + b^2}{8a^2b^2} = 1 \implies a^2 + b^2 = 8a^2b^2,
$
$
\therefore \dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{a^2 + b^2}{a^2b^2} = 8.
$

故选：D.