Question

If a hyperbola has one focus at origin and its eccentricity is√2 one of the directrices is x+y+1=0 Then find the centre of the hyperbola and the equation of its asymptotes

Answer 1

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Answer 2

Answer:

The centre of this hyperbola is $(-1,-1)$.

The equation of the asymptotes are $x+1=0 , y+1=0$

Step-by-step explanation:

We know that :

eccentricity $=\frac{distance of point on hyperbola to force}{distance of point on hyperbola to directrix}$

Let us consider a general point $(x,y)$ on the hyperbola.

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

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

$2x+2y+2xy+1=0$  -- This is the equation of the conic

$\frac{d(c)}{dx} =2+2y$ --- $c=$ equation of conic

Therefore, $\frac{dc}{dx} =0$ ; $2+2y=0$

Hence, $y=-1$

Also, $\frac{d(c)}{dy} =2+2x=0$

Hence, $x=-1$

Therefore the centre of hyperbola (conic) is $(-1,-1)$

To find the equation of its asymptotes :

We know that the standard form is : $S=\frac{x^2}{a^2} -\frac{y^2}{b^2} =1$

Focus : $(0,0)$

Eccentricity : $\sqrt{2}$

Equation of the directrix : $x+y+1=0$

We know that for any point on S, say $P(h,k)$ :

Distance of the point $P$ from focus, $FP$ $= e*$ (perpendicular distance of the point p from the directrix)

In this case,

$FP=\sqrt{h^2+k^2} =e*\frac{h+k+1}{1^2+1^2}$

On squaring both sides, we get :

$h^2+k^2= \sqrt{2} ^2\frac{(h+k+1)^{2} }{\sqrt{2} ^2}$

$h^2+k^2=h^2+k^2+1+2hk+2h+2k\\2hk+2h+2k+1=0$

From the general equation $2xy+2x+2y+1=c$

On differentiating $C$ with respect to $x$ and $y$, we get the asymptote's equation.

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

Similarly,

$\frac{dc}{dy} =2x+0+1+0=0\\x+1=0$

Hence the equation of the asymptotes are $x+1=0 , y+1=0$