Let $a, b$ be two known constants, and $a>b$. Points $P^{3}, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, then $\frac{|A Q| \cdot |A R|}{|O P|^{2}}=$ $\qquad$ (where $O$ is the origin).
11. 2.

Let the coordinates of point $P$ be $P(a \cos \theta, b \sin \theta)$, then $k_{\left(P^{P}\right.}=\frac{b}{a} \tan \theta$. Draw $P P_{1} \perp x$-axis at point $P_{1}$, and $Q Q_{1} \perp x$-axis at point $Q_{1}$, then $\left|O P_{1}\right|=|a \cos \theta|$.
Since $A R \parallel O P$, we have $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=\frac{\left|A Q_{1}\right| \cdot|A O|}{\left|O P_{1}\right|^{2}}$.
The equation of $A R$ is $y=\frac{b}{a} \tan \theta(x+a)$, substituting into the ellipse equation, eliminating $y$, we get
$$
\left(1+\tan ^{2} \theta\right) x^{2}+2 a \tan ^{2} \theta x+\left(\tan ^{2} \theta-1\right) a^{2}=0 .
$$

One of its roots is $-a$, the other root $x_{1}$ is the x-coordinate of point $Q$, by Vieta's theorem we know $-a \cdot x_{1}=\frac{\tan ^{2} \theta-1}{1+\tan ^{2} \theta} a^{2}, x_{1}=$
$$
\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} a, \left|A Q_{1}\right|=a+x_{1}=\frac{2 a}{1+\tan ^{2} \theta}=2 a \cos ^{2} \theta, \text{ hence } \frac{|A Q| \cdot|A R|}{|O P|^{2}}=\frac{2 a \cos ^{2} \theta \cdot a}{a^{2} \cos ^{2} \theta}=2 \text{. }
$$