If 5x + 9 = 0 is the directrix of the hyperbo | vedclass

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(C) The given equation of the hyperbola is $16x^2 - 9y^2 = 144.$ Dividing by $144,$ we get $\frac{x^2}{9} - \frac{y^2}{16} = 1.$Here,$a^2 = 9$ and $b^

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$(-5, 0)$

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(C) The given equation of the hyperbola is $16x^2 - 9y^2 = 144.$ Dividing by $144,$ we get $\frac{x^2}{9} - \frac{y^2}{16} = 1.$Here,$a^2 = 9$ and $b^2 = 16,$ so $a = 3$ and $b = 4.$The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}.$The directrix of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x = \pm \frac{a}{e}.$Given directrix is $5x + 9 = 0,$ which is $x = -\frac{9}{5}.$Since $\frac{a}{e} = \frac{3}{5/3} = \frac{9}{5},$ the directrix $x = -\frac{9}{5}$ corresponds to the focus at $(-ae, 0).$Thus,the focus is $(-3 	imes \frac{5}{3}, 0) = (-5, 0).$

If $5x + 9 = 0$ is the directrix of the hyperbola $16x^2 - 9y^2 = 144,$ then its corresponding focus is

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