If 5x + 9 = 0 is the directrix of the hyperbola 16 x ^ 2 - 9 y ^ 2...

The equation of a hyperbola in standard form is $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$ and its foci are \left(\pm ae, 0\right)$and directrices are$x = \pm \dfrac{a}{e} .$</p><p>The given equation of hyperbola is$16{x}^{2} - 9{y}^{2} = 144, ⇒ \dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{16} = 1$</p><p>$⇒ {a}^{2} = 9, {b}^{2} = 16$</p><p>Given, directrix of the hyperbola is$5x + 9 = 0, ⇒ x = - \dfrac{9}{5} .$</p><p>Hence, on comparing the directrix with the standard equation, we get$\dfrac{a}{e} = \dfrac{9}{5}$</p><p>$⇒ \dfrac{3}{e} = \dfrac{9}{5}$</p><p>$⇒ e = \dfrac{5}{3}$</p><p>And, hence the required focus is \left(- ae, 0\right) = ( - 5, 0 ) .$