Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the vertices are $\left(\pm 5, 0\right)$.
Let the foci be $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$
$\therefore$ $\;$ The foci $F_1$ and $F_2$ lie on the the X axis.
Let the equation of the ellipse be $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
Let the vertices be $V_1 = \left(5,0\right)$ and $V_2 = \left(-5,0\right)$
Now, $\text{length } \left(V_1 V_2\right) = 2a$
i.e. $\;$ $\sqrt{\left(5 + 5\right)^2} = 2a$
i.e. $\;$ $10 = 2a$ $\implies$ $a = 5$
and $\text{length } \left(F_1 F_2\right) = 2 a e$
i.e. $\;$ $\sqrt{\left(3 + 3\right)^2} = 2 \times 5 \times e$
i.e. $\;$ $6 = 10 e$ $\implies$ $e = \dfrac{3}{5}$
Also, $\;$ $b^2 = a^2 \left(1 - e^2\right)$
i.e. $\;$ $b^2 = \left(5\right)^2 \times \left[1 - \left(\dfrac{3}{5}\right)^2\right]$
i.e. $\;$ $b^2 = 25 \times \left[\dfrac{16}{25}\right]$ $\implies$ $b^2 = 16$
$\therefore$ $\;$ Required equation of ellipse is
$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$
i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$