Question

The equation of the directrix of a hyperbola is x-y+3 =0.Its ficys us(-1,1) and eccentricity 3.Find the equation of the hyperbola.

Answer 1

$\text{Let S(-1,1) be the focus nd P(x,y) be a point on the hyperbola Draw PM perpenicular from p on the directrix.Then,by definition.}SP=ePM\Rightarrow S{P}^2=e^{2}P{M}^2\Rightarrow (x-1{)}^2+(Y-1{)}^2=\left(3{)}^2{\left[\begin{array}{c}\frac{x-y+3}{\sqrt{1^1+(-1{)}^2}}\end{array}\right]}^2\right[\because e=3]\Rightarrow x^2+1+2x+y^2+1-2y={\frac{9[x-y+3]}{2}}^2\Rightarrow 2[x^2+y^2+2x-2y+2]=9[x-y+3{]}^2\Rightarrow 2x^2+2y^2+4x-4y+4=9\left[x^{2}9\right(-y{)}^2+3^2+2\times x\times (-y)\times 3+2\times 3\times x]\Rightarrow 2x^2+2y^2+4x-4y-4=9[x^2+y^2+9-2xy-6y+6x]\Rightarrow 2x^2+2y^2+4x-4y+4=9x^2+9y^2+81-18xy-54y+4y+81-4=0\Rightarrow 7x^2+7y^2-18xy+50x-50y+77=0\text{This is the required equation of the hyperbola.}$