5 Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ 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 $R, O$ being the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\qquad$.
5 2. Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter)
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$

Therefore,
$$
|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$

In (1), let $x=0$, we get $t=\frac{\alpha}{\cos \theta}$. So $|A R|=\frac{\alpha}{|\cos \theta|}$.

Also, let
$$
O P:\left\{\begin{array}{l}
x=t^{\prime} \cos \theta \\
y=t^{\prime} \sin \theta
\end{array} \text { ( } t^{\prime} \text { is a parameter }\right)
$$

Substitute (2) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
t^{\prime 2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$

Therefore,
$$
|O P|^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$

Hence,
$$
\frac{|A Q| \cdot|A R|}{|O P|^{2}}=\cdots=2 .
$$