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**Hint:**First apply the basic definition of hyperbola to get the equation of hyperbola. One of the focus is given, find the other focus then find the midpoint of both the focus that will be the centre of the hyperbola. To find the equation of a directrix use the concept of distance between two parallel lines and equate it with the distance.

**Complete step by step solution:**Let P(x,y) be any point on the hyperbola

and l1 is perpendicular from P on the directrix l1.

Then by definition

\[{F_{1}}P = \dfrac{c}{q} = eP{l_1}\]

Squaring both side

\[{\left( {{F_{1}}P} \right)^2} = {\left( {eP{l_1}} \right)^2}\]

Or,

\[\;{\left( {x - a} \right)^2} + {\left( {y - 0} \right)^2} = \dfrac{{25}}{{16}} \times {\left( {\dfrac{{4x - 3y - a}}{{\sqrt {16 + 9} }}} \right)^2}\]

\[16{x^2} + 16{y^2} + 16{a^2} - 32 ax = 16{x^2} + 9{y^2} - 24xy + {a^2} - 8ax + 6ay\]

On cancelling the like terms we get the equation of hyperbola as

\[7{y^2} + 15{a^2} + 24xy - 24ax - 6ay = 0\]

Distance of l1 from F1

\[ = \dfrac{{4\left( a \right) - 3\left( 0 \right) - a}}{{\sqrt {{4^2} + {3^2}} }}\]

\[\begin{gathered}

= \dfrac{{3a}}{5} \\

= \dfrac{{{c^2} - {q^2}}}{c} \\

\end{gathered} \]

We know

\[\dfrac{c}{q} = e\;\]

\[c = \dfrac{{3a{e^{2}}}}{{({e^2} - 1)5}}\]

\[c = \dfrac{{5a}}{3}\]

$F_1F_2$ =2c and $F_1F_2$ is perpendicular to l1

∴ F1F2 is at a distance of 2c at slope = \[\dfrac{{ - 3}}{4}\] towards l1

on solving we get \[{F_2} = (a - 2 \times \dfrac{{4c}}{5},0 - 2 \times \dfrac{{3c}}{5})\]

\[{F_2} = (\dfrac{{ - 5a}}{3},2a)\]

Centre of the hyperbola is the midpoint of F1F2 ie. at \[(\dfrac{{ - a}}{3},a)\]

Let the equation of directrix l2 be 4x−3y=k ∵l1 and l2 are parallel with distance between them equal to \[\dfrac{{2{q^{2}}}}{c} = \dfrac{{32c}}{{25}}\]

distance between 2 parallel lines \[Ax + By + {C_{1}} = 0\;\] & \[Ax + By + {C_{2}} = 0\;\] is given by \[\dfrac{{\mid {C_2} - {C_1}\mid \;}}{{\sqrt {{A^2} + {B^2}} }}\]

\[\begin{array}{*{20}{l}}

{\dfrac{{k - a}}{5} = \pm \dfrac{{32a}}{{15}}} \\

{k = a \pm \dfrac{{32a}}{3}}

\end{array}\]

We get 2 values of k, one value for above l2 and other for below, we see that the value required is k for line above l1∵l2 is in direction of centre relative to l1 and centre>yF1

or you can check by drawing graph

∴\[k = - \dfrac{{ - 29a}}{3}\]

Equation of directrix is \[12x - 9y + 29a = 0\]

Centre of hyperbola is at \[(\dfrac{{ - a}}{3},a)\]

Equation of hyperbola is \[7{y^2} + 15{a^2} + 24xy - 24ax - 6ay = 0\]

**Note:**Generally equations of any type of conics can be found by using the basic definition of conic. So students are always aware of basic conic conditions.

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