If the two ends of the major axis of an ellipse are $(5, 0)$ and $(-5, 0)$ and one focus lies on the line $3x - 5y - 9 = 0$, then its equation is

The ends of the major axis are given as $(5, 0)$ and $(-5, 0)$.

This implies that the center of the ellipse is at the origin $(0, 0)$ and the semi-major axis is $a = 5$.

The foci of the ellipse lie on the major axis, which is the x-axis. Thus, the coordinates of the foci are of the form $(\pm ae, 0)$.

It is given that one focus lies on the line $3x - 5y - 9 = 0$. Substituting $y = 0$ into the equation of the line:

$3x - 5(0) - 9 = 0 \Rightarrow 3x = 9 \Rightarrow x = 3$

So, one focus is at $(3, 0)$, which means $ae = 3$.

Using the relation $b^2 = a^2(1 - e^2) = a^2 - (ae)^2$, we get:

$b^2 = 5^2 - 3^2 = 25 - 9 = 16$

The equation of the ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Substituting the values of $a^2$ and $b^2$, we obtain:

$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

Answer: $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$