Question

Find the equation to the hyperbola, whose eccentricity is 45​, whose focus is (a,0), and whose directrix is 4x−3y=a. Find also the coordinates of the centre and the equation to the other directrix.

Answer 1

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
\[\dfrac{c}{q} = e\;\]
\[c = \dfrac{{3a{e^{2}}}}{{({e^2} - 1)5}}\]
\[c = \dfrac{{5a}}{3}\]
$F_1​F_2$ ​=2c and $F_1​F_2$ is perpendicular to l1​
∴ F1​F2​ 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 F1​F2​ 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.