Step 1: The standard form of the ellipse with major axis on the x-axis is:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Step 2: Since the point $(4, 3)$ lies on the ellipse, substituting gives:

$$\frac{16}{a^2} + \frac{9}{b^2} = 1 \quad \ldots \text{(1)}$$

Step 3: Since the point $(6, 2)$ lies on the ellipse, substituting gives:

$$\frac{36}{a^2} + \frac{4}{b^2} = 1 \quad \ldots \text{(2)}$$

Step 4: Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. Then equations become:

$$16u + 9v = 1 \quad \text{and} \quad 36u + 4v = 1$$

Step 5: Multiply first equation by 4:

$$64u + 36v = 4$$

Multiply second equation by 9:

$$324u + 36v = 9$$

Step 6: Subtracting:

$$260u = 5$$

so $u = \frac{1}{52}$. Therefore, $a^2 = 52$.

Step 7: Substituting in equation (1):

$$\frac{16}{52} + \frac{9}{b^2} = 1$$

which gives:

$$\frac{9}{b^2} = \frac{9}{13}$$

Therefore, $b^2 = 13$.

Step 8: The required equation is:

$$\frac{x^2}{52} + \frac{y^2}{13} = 1$$

or:

$$x^2 + 4y^2 = 52$$

∴ The equation of the ellipse is $\mathbf{x^2 + 4y^2 = 52}$.