(p, q) is the point of intersection of a latu | vedclass

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

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$\frac{15}{4}$

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(C) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$. Dividing by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Here,$a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$. The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. The equation of the latus rectum is $x = ae = 4 	imes \frac{5}{4} = 5$. The equation of the asymptote is $\frac{x^2}{16} - \frac{y^2}{9} = 0$,which simplifies to $y = \pm \frac{b}{a}x = \pm \frac{3}{4}x$. For the first quadrant,we take $y = \frac{3}{4}x$. Substituting $x = 5$ into the asymptote equation,we get $q = \frac{3}{4}(5) = \frac{15}{4}$. Thus,$q = \frac{15}{4}$.

$(p, q)$ is the point of intersection of a latus rectum and an asymptote of the hyperbola $9x^2 - 16y^2 = 144$. If $p > 0$ and $q > 0$,then $q =$

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