Question

Find the equation of hyperbola whose focus is $(2,-1)$, directrix is $2x+3y=1$ and eccentricity $=2$

Answer 1

Given that: $e=2,S(2,-1)$ and equation of directrix is $2x+3y-1=0$

Let a point be $P(x,y)$, such that
$SP=ePM$, where $PM$ is perpendicular distance from $P(x,y)$ to directrix

$\Rightarrow \sqrt{(x-2{)}^2+(y+1{)}^2}=2\times \left|\frac{2x+3y-1}{\sqrt{2^2+3^2}}\right|$

$\Rightarrow \sqrt{(x-2{)}^2+(y+1{)}^2}=\frac{2}{\sqrt{13}}|2x+3y-1|$

On squaring both sides, we get
$\Rightarrow (x-2{)}^2+(y+1{)}^2=\frac{4}{13}(2x+3y-1{)}^2$
$\Rightarrow (x-2{)}^2+(y+1{)}^2=4/13(2x+3y-1{)}^2$
$\Rightarrow 13(x^2-4x+4+y^2+2y+1)=4(4x^2+9y^2+1+12xy-6y-4x)$
$\Rightarrow 13x^2-52x+13y^2+26y+65=16x^2+36y^2+4+48xy-24y-16x$
$\Rightarrow 3x^2+23y^2+48xy+36x-50y-61=0$

$\therefore$ Equation of hyperbola is
$3x^2+48xy+23y^2+36x-50y-61=0$