Question

Find the equation of hyperbola whose focus is $(1,1)$, directrix is $3x+4y+8=0$ and eccentricity $=2$

Answer 1

Given that : $e=2,S(1,1)$ and equation of directrix is $3x+4y+8=0$

Let a point be $P(x,y)$, such that

$SP=ePM$, where $PM$ is perpendicular distance from $P(x,y)$ to directrix

$\Rightarrow \sqrt{(x-1{)}^2+(y-1{)}^2}=2\times \left|\frac{3x+4y+8}{\sqrt{3^2+4^2}}\right|$

$\Rightarrow \sqrt{(x-1{)}^2+(y-1{)}^2}=\frac{2}{5}|3x+4y+8|$

Squaring both sides, we get
$\Rightarrow (x-1{)}^2+(y-1{)}^2=\frac{4}{25}(3x+4y+8{)}^2$
$\Rightarrow 25(x^2-2x+1+y^2-2y+1)=4(9x^2+16y^2+64+24xy+64y+48x)$
$\Rightarrow 25x^2-50x+25y^2-50y+50=36x^2+64y^2+256+96xy+256y+192x$
$\Rightarrow 11x^2+39y^2+96xy+242x+306y+206=0$

$\therefore$ Equation of hyperbola is
$11x^2+96xy+39y^2+242x+306y+206=0$