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Q.
The equation of the conic with focus at $(1,-1)$ directrix along $x - y +1=0$ and with eccentricity $\sqrt{2}$ is
Conic Sections
Solution:
From the definition of conic; If $P(x, y)$ is the point on a conic then ratio of its distance from focus to its distance from directrix is a fixed ratio e, called eccentricity. Here focus is $(1,-1)$ and directrix is $x-y+1=0$.
Distance of this point from focus $=\sqrt{( x -1)^{2}+( y +1)^{2}}$
Distance of this point from directrix. $=\left|\frac{ x - y +1}{\sqrt{1^{2}+(-1)^{2}}}\right|$
So, from the definition of conic
$\sqrt{(x-1)^{2}+(y+1)^{2}}=e .\left|\frac{x-y+1}{\sqrt{2}}\right| \,\,\,\,\,\,\,\,\,...(i)$
Squaring both sides of equation (i), we get
$(x-1)^{2}+(y+1)^{2}=e^{2} \cdot \frac{(x-y+1)^{2}}{2}=(\sqrt{2})^{2} \frac{(x-y+1)^{2}}{2}$
$=(x-y+1)^{2}$
$\Rightarrow (x-1)^{2}+(y+1)^{2}=(x-y+1)^{2}$
$\Rightarrow x^{2}-2 x+1+y^{2}+2 y+1=x^{2}-2 x y+y^{2}+2 x-2 y+1$
$\Rightarrow 2 x y-4 x+4 y+1=0$