The problem can be solved by applying the no-slip condition at the contact point of the two gears and using the geometric properties of an ellipse.

1. Kinematic Condition Let $r_1$ and $r_2$ be the distances from the rotation centers $O_1$ and $O_2$ to the contact point $P$, respectively. Since the gears are locked in motion, there is no slipping at the contact point. The tangential velocities of the two gears at point $P$ must be equal.

   $$v_1 = v_2$$

   As the contact point lies on the line $O_1O_2$, the velocities are perpendicular to this line. Their magnitudes are given by:

   $$\omega_1 r_1 = \omega_2 r_2$$

   This gives the ratio of the angular velocities:

   $$\frac{\omega_2}{\omega_1} = \frac{r_1}{r_2}$$

2. Geometric Constraints The problem states that the distance between the centers of rotation is $O_1O_2 = 2a$. Since the contact point $P$ lies on the line segment $O_1O_2$, the sum of the distances from the foci to the contact point is:

   $$r_1 + r_2 = 2a \quad \implies \quad r_2 = 2a - r_1$$

   Substituting this into the kinematic relation gives $\omega_2$ in terms of $r_1$:

   $$\omega_2 = \omega_1 \frac{r_1}{2a - r_1}$$

3. Ellipse Polar Equation The shape of each gear is an ellipse. The distance from a focus to a point on the ellipse can be described by the polar equation. With the origin at the focus $O_1$, the distance $r_1$ to a point on the perimeter is given by:

   $$r_1(\phi) = \frac{a(1-e^2)}{1 - e \cos\phi}$$

   where $\phi$ is the angle between the major axis and the line segment $O_1P$ (which is the line $O_1O_2$), and $e$ is the eccentricity of the ellipse. The eccentricity is defined as $e = c/a = \sqrt{a^2-b^2}/a$.

4. Derivation of $\omega_2$ Substitute the expression for $r_1(\phi)$ into the equation for $\omega_2$:

   $$\omega_2 = \omega_1 \frac{\frac{a(1-e^2)}{1 - e \cos\phi}}{2a - \frac{a(1-e^2)}{1 - e \cos\phi}}$$

   Simplify the expression:

   $$\omega_2 = \omega_1 \frac{a(1-e^2)}{2a(1 - e \cos\phi) - a(1-e^2)} = \omega_1 \frac{a(1-e^2)}{2a - 2ae \cos\phi - a + ae^2}$$ $$\omega_2 = \omega_1 \frac{a(1-e^2)}{a + ae^2 - 2ae \cos\phi} = \omega_1 \frac{1-e^2}{1 + e^2 - 2e \cos\phi}$$

   To express this in terms of the given parameters $a$ and $b$, we use $e^2 = \frac{a^2-b^2}{a^2}$:

   $$1 - e^2 = 1 - \frac{a^2-b^2}{a^2} = \frac{b^2}{a^2}$$ $$1 + e^2 = 1 + \frac{a^2-b^2}{a^2} = \frac{2a^2-b^2}{a^2}$$

   Substituting these into the expression for $\omega_2$:

   $$\omega_2 = \omega_1 \frac{b^2/a^2}{\frac{2a^2-b^2}{a^2} - 2\frac{\sqrt{a^2-b^2}}{a}\cos\phi}$$ $$\omega_2 = \omega_1 \frac{b^2}{2a^2-b^2 - 2a\sqrt{a^2-b^2}\cos\phi}$$