Question

Find the equation of the hyperbola whose eccentricity is 2, one focus is $(1,1)$ and the corresponding directrix is $(x+y+1)=0.$

Answer 1

Let $S(1,1)$ be focus and $P(x,y)$ be the point on the hyperbola.

From $P$ draw $PM$ perpendicular to the directrix then $PM=\frac{x+y+1}{\sqrt{1^2+1^2}}=\frac{x+y+1}{\sqrt{2}}$

By the definition of hyperbola we have $\frac{SP}{PM}=e$

$\Rightarrow SP=ePM$

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

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

$\Rightarrow (x-1{)}^2+(y-1{)}^2=2(x+y+1{)}^2$

$\Rightarrow x^2-2x+1+y^2-2y+1=2(x^2+y^2+1+2xy+2x+2y)$

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

$\Rightarrow x^2+y^2+4xy+6x+6y=0$