Example 6 Given that the slope of line $l$ is $\frac{1}{2}$, and it passes through the left focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, intersecting the ellipse at points $A, B$. The center of the ellipse is $O$, and the distance from point $O$ to line $A B$ is $d=1$. The length of chord $A B$ is $\frac{4}{5}$ of the length of the major axis of the ellipse. Find the equation of the ellipse.
From the problem, we can set the equation of $AB$ as $y=\frac{1}{2}(x+c)$, and the distance from it to the origin $d=\frac{c}{\sqrt{5}}$, i.e., $\frac{c}{\sqrt{5}}=1$, hence $c^{2}=5$.

Also, $|AB|=\frac{4}{5} \cdot 2a$, by property 11 (III), we have $\frac{2}{a} \cdot \frac{8a}{5}+\frac{1}{|OP|^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$, thus, we get $\frac{1}{|OP|^{2}}=$ $\frac{1}{b^{2}}-\frac{1}{4a^{2}}$.

It is also easy to know that the equation of $OP$ is $y=-2x$, substituting it into the ellipse equation, we solve to get $x^{2}=\frac{a^{2}b^{2}}{4a^{2}+b^{2}}, y^{2}=$ $\frac{4a^{2}b^{2}}{4a^{2}+b^{2}}$. Therefore, $|OP|^{2}=x^{2}+y^{2}=\frac{5a^{2}b^{2}}{4a^{2}+b^{2}}$, and substituting it into (*) equation and simplifying, we get $4a^{2}=9b^{2}$. Considering $c^{2}=$ 5, we find $a=3, b=2$. Hence, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$.