Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $9x^2 + 16y^2 = 144$
Equation of the given ellipse is: $\;\;$ $9x^2 + 16y^2 = 144$
i.e. $\;$ $\dfrac{x^2}{144 / 9} + \dfrac{y^2}{144 / 16} = 1$
i.e. $\;$ $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 16 \implies a = 4$; $\;$ $b^2 = 9$
Now, $\;$ $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{9}{16} = \dfrac{7}{16}$ $\implies$ $e = \dfrac{\sqrt{7}}{4}$
Equations of directrices are: $\;$ $x = \pm \dfrac{a}{e}$
i.e. $\;$ $x = \pm \dfrac{4}{\sqrt{7} / 4}$
i.e. $\;$ $x = \pm \dfrac{16}{\sqrt{7}}$
Equations of latus rectum are: $\;$ $x = \pm ae$
i.e. $\;$ $x = \pm 4 \times \dfrac{\sqrt{7}}{4}$
i.e. $\;$ $x = \pm \sqrt{7}$
Length of latus rectum $= \dfrac{2b^2}{a} = 2 \times \dfrac{9}{4} = \dfrac{9}{2}$