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Let $(1,2)$ be the focus and $x+y+1=0$ be the directrix of a hyperbola $\mathrm{H}$. If $\sqrt{3}$ is the eccentricity of $\mathrm{H}$, then its equation is
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The correct answer is:
$x^2+6 x y+y^2+10 x+14 y-7=0$
Given equation of directrix is $x+y+1=0$ and Focus is $(1,2)=5$
Let $P(x, y)$ be any point on hyperbola $\mathrm{H}$
Let $\mathrm{PM}$ be the length of the perpendicular from $\mathrm{P}$ to the directrix.
then
$\frac{P S}{P M}=\sqrt{3} \Rightarrow P S^2=3 P M^2$
$\begin{aligned} & \Rightarrow(x-1)^2+(y-2)^2=3\left|\frac{x+y+1}{\sqrt{2}}\right|^2 \\ & \begin{aligned} \Rightarrow 2\left(x^2+1-2 x+y^2+4-4 y\right)=3\left(x^2+y^2+1\right. & +2 x y \\ & +2 y+2 x) \\ \Rightarrow 2 x^2+2-4 x+2 y^2+8-8 y=3 x^2+3 y^2+3 & +6 x y +6 y+6 x\\ \Rightarrow x^2+y^2+6 x y+10 x+14 y-7=0 &\end{aligned}\end{aligned}$
Let $P(x, y)$ be any point on hyperbola $\mathrm{H}$
Let $\mathrm{PM}$ be the length of the perpendicular from $\mathrm{P}$ to the directrix.
then
$\frac{P S}{P M}=\sqrt{3} \Rightarrow P S^2=3 P M^2$
$\begin{aligned} & \Rightarrow(x-1)^2+(y-2)^2=3\left|\frac{x+y+1}{\sqrt{2}}\right|^2 \\ & \begin{aligned} \Rightarrow 2\left(x^2+1-2 x+y^2+4-4 y\right)=3\left(x^2+y^2+1\right. & +2 x y \\ & +2 y+2 x) \\ \Rightarrow 2 x^2+2-4 x+2 y^2+8-8 y=3 x^2+3 y^2+3 & +6 x y +6 y+6 x\\ \Rightarrow x^2+y^2+6 x y+10 x+14 y-7=0 &\end{aligned}\end{aligned}$
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