22. Solution: (1) Since $-1 < \frac{1-t^2}{1+t^2} \leq 1$ and $x^2 + \left(\frac{y}{2}\right)^2 = \left(\frac{1-t^2}{1+t^2}\right)^2 + \frac{4t^2}{(1+t^2)^2} = 1$, the rectangular coordinate equation of $C$ is $x^2 + \frac{y^2}{4} = 1$ $(x \neq -1)$. The rectangular coordinate equation of $l$ is $2x + \sqrt{3}y + 11 = 0$. (2) From (1) we can set the parametric equation of $C$ as $\left\{\begin{array}{l} x = \cos\alpha, \\ y = 2\sin\alpha \end{array}\right.$ ($\alpha$ is the parameter, $-\pi < \alpha < \pi$). The distance from a point on $C$ to $l$ is $\frac{|2\cos\alpha + 2\sqrt{3}\sin\alpha + 11|}{\sqrt{7}} = \frac{4\cos\left(\alpha - \frac{\pi}{3}\right) + 11}{\sqrt{7}}$. When $\alpha = -\frac{2\pi}{3}$, $4\cos\left(\alpha - \frac{\pi}{3}\right) + 11$ attains its minimum value of 7, therefore the minimum distance from a point on $C$ to $l$ is $\sqrt{7}$.
22. Solution: (1) Since $-1 < \frac{1-t^2}{1+t^2} \leq 1$ and $x^2 + \left(\frac{y}{2}\right)^2 = \left(\frac{1-t^2}{1+t^2}\right)^2 + \frac{4t^2}{(1+t^2)^2} = 1$, the rectangular coordinate equation of $C$ is $x^2 + \frac{y^2}{4} = 1$ $(x \neq -1)$.\\
The rectangular coordinate equation of $l$ is $2x + \sqrt{3}y + 11 = 0$.\\
(2) From (1) we can set the parametric equation of $C$ as $\left\{\begin{array}{l} x = \cos\alpha, \\ y = 2\sin\alpha \end{array}\right.$ ($\alpha$ is the parameter, $-\pi < \alpha < \pi$).\\
The distance from a point on $C$ to $l$ is $\frac{|2\cos\alpha + 2\sqrt{3}\sin\alpha + 11|}{\sqrt{7}} = \frac{4\cos\left(\alpha - \frac{\pi}{3}\right) + 11}{\sqrt{7}}$.
When $\alpha = -\frac{2\pi}{3}$, $4\cos\left(\alpha - \frac{\pi}{3}\right) + 11$ attains its minimum value of 7, therefore the minimum distance from a point on $C$ to $l$ is $\sqrt{7}$.