If one focus of a hyperbola is (3,0),the equa | vedclass

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(B) The distance between the focus and the directrix is given by $d = \frac{a}{e} - ae = a(\frac{1}{e} - e)$ (for hyperbola,distance is $a/e - ae$ or

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$\left(\frac{11}{5}, \frac{3}{5}\right)$

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(B) The distance between the focus and the directrix is given by $d = \frac{a}{e} - ae = a(\frac{1}{e} - e)$ (for hyperbola,distance is $a/e - ae$ or $ae - a/e$ depending on orientation). Given $e = 5/4$,the distance is $a(5/4 - 4/5) = a(9/20)$.The perpendicular distance from the focus $(3,0)$ to the directrix $4x - 3y - 3 = 0$ is $\frac{|4(3) - 3(0) - 3|}{\sqrt{4^2 + (-3)^2}} = \frac{9}{5}$.Equating the two: $\frac{9a}{20} = \frac{9}{5} \implies a = 4$.The axis of the hyperbola is perpendicular to the directrix. Since the directrix slope is $4/3$,the axis slope is $-3/4$.The unit vector along the axis towards the vertex is $(-\frac{4}{5}, \frac{3}{5})$.The distance from the focus to the vertex is $ae - a = 4(5/4) - 4 = 1$.Thus,the vertex is $(3, 0) + 1 	imes (-\frac{4}{5}, \frac{3}{5}) = (3 - \frac{4}{5}, 0 + \frac{3}{5}) = (\frac{11}{5}, \frac{3}{5})$.

If one focus of a hyperbola is $(3,0)$,the equation of its directrix is $4x - 3y - 3 = 0$ and its eccentricity is $e = \frac{5}{4}$,then the coordinates of its vertex are:

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