Question

Find the equation to the hyperbola whose directrix is $2x+y=1$, focus $(1,1)$ & eccentricity $\sqrt{3}$. Find also the length of its latus rectum.

Answer 1

Solution :-

We have to find equation of hyperbola, given directrix $2x+y=1,$

foucas $(1,1)$ & eccentricity $\sqrt{3}$

Let $P(x,y)$ be any point on the hyperbola Drow PM $\perp$ from

Pm the directrix . Then

SP = ePM

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

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

$\Rightarrow 3x^2+5y^2-10x-10y+10=12x^2+3y^2+3+12xy-6y-12x$

$\Rightarrow 7x^2-2y^2+12xy-2x+4y-7=0$