Question

If a hyperbola has one focus at the origin in and its eccentricity is $\sqrt{2}$. One of the directrices is $x+y+1=0$. The the centre of the hyperbola is.

• A. $(-1,-1)$
• B. $(1,-1)$
• C. $(-2,-1)$
• D. $(2,2)$

Answer 1

The correct option is A $(-1,-1)$

We know that

$eccentricity=\frac{distanceofpointonhyperbolatoforce}{distanceofpointonhyperbolatodirectrix}$

$\therefore$ Let a general point $(x,y)$ on hyperbola.

$\therefore \sqrt{2}=\frac{\sqrt{(x-0{)}^2+(y-0{)}^2}}{\frac{x+y+1}{\sqrt{2}}}$

$\therefore \sqrt{2}\frac{(x+y+1)}{\sqrt{2}}=\sqrt{x^2+y^2}$

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

$\therefore x^2+y^2+1+2xy+2y+2x=x^2+y^2$

$\therefore 2x+2y+2xy+1=0\to Equationofconic$

$\therefore \frac{d\left(c\right)}{dx}=2+2y\to c=Equationofconic$

$\therefore \frac{dc}{dx}=0;2+2y=0;y=-1$

Also $\frac{d\left(c\right)}{dy}=2+2x=0;x=-1$

Centre of conic (hyperbola)is $(-1,-1)$