Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the length of the latus rectum is $\dfrac{32}{5}$
Let the foci be $\;$ $F_1 = \left(3,0\right)$; $\;\;$ $F_2 = \left(-3,0\right)$
Since the foci lie on the X axis, the major axis of the ellipse is along the X axis.
$\therefore$ $\;$ Let the equation of the required ellipse be
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;\;\; \cdots \; (1)$
Now, length$\left(F_1 F_2\right) = 2ae$ $\;\;\; \cdots \; (2)$
length of latus rectum $= \dfrac{2b^2}{a}$ $\;\;\; \cdots \; (3)$
and $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (4)$
Given: $\;\;$ Length of latus rectum $= \dfrac{32}{5}$
$\implies$ $\dfrac{2b^2}{a} = \dfrac{32}{5}$ $\implies$ $b^2 = \dfrac{16a}{5}$ $\;\;\; \cdots \; (5a)$
Also, $\;$ $F_1 F_2 = \sqrt{\left(3 + 3\right)^2} = 6$
$\therefore$ $\;$ We have $\;\;$ $6 = 2ae$ $\implies$ $e = \dfrac{3}{a}$ $\;\;\; \cdots \; (5b)$
$\therefore$ $\;$ We have from equations $(4)$, $(5a)$ and $(5b)$,
$\dfrac{16a}{5} = a^2 \left(1 - \dfrac{9}{a^2}\right)$
i.e. $\;$ $\dfrac{16 a}{5} = a^2 - 9$
i.e. $\;$ $5a^2 - 16a - 45 = 0$
i.e. $\;$ $\left(a - 5\right) \left(5a + 9\right) = 0$
i.e. $\;$ $a = 5$ $\;\;\; \cdots \; (6)$ $\;$ or $\;$ $a = \dfrac{-9}{5}$ $\;$ (not an acceptable solution)
$\therefore$ $\;$ In view of equation $(6)$, we have from equation $(5a)$
$b^2 = \dfrac{16 \times 5}{5} = 16$
$\therefore$ $\;$ Substituting the values of $a$ and $b^2$ in equation $(1)$ gives the equation of the required ellipse as
$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$
i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$