Find the equation of the hyperbola whose one directrix is $2 x + y = 1$ , the corresponding focus is $(1, 2)$ and eccentricity is $\sqrt{3}$

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Solution

Equation of directrix is $2 x + y - 1 = 0$ .
Let $S$ be the corresponding focus, then $S = (1, 2)$ .
Given $e = \sqrt{3}$
Let $P (x, y)$ be any point on the required hyperbola.
Let $P M$ be the length of the perpendicular from $P$ to the directrix.
Then, $\frac{P S}{P M} = e = \sqrt{3}$
$∴ {P S}^{2} = 3 {P M}^{2}$
or ${(x - 1)}^{2} + {(y - 2)}^{2} = 3 {∣ ∣ ∣ \frac{2 x + y - 1}{\sqrt{5}} ∣ ∣ ∣}^{2}$
or $5 [{(x - 1)}^{2} + {(y - 2)}^{2}] = 3 (4 x^{2} + y^{2} + 1 + 4 x y - 2 y - 4 x)$
or $5 [x^{2} + 1 - 2 x + y^{2} - 4 y + 4] = 12 x^{2} + 3 y^{2} + 3 + 12 x y - 6 y - 12 x$
$\Rightarrow 5 x^{2} + 5 - 10 x + 5 y^{2} - 20 y + 20 - 12 x^{2} - 3 y^{2} - 3 - 12 x y + 6 y + 12 x = 0$
$\Rightarrow - 7 x^{2} + 2 y^{2} - 12 x y + 2 x - 14 y + 22 = 0$
or $7 x^{2} - 2 y^{2} + 12 x y - 2 x + 14 y - 22 = 0$

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