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

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Solution

Given that : $e = \sqrt{3}, S (1, 1)$ and equation of directrix is $2 x + y - 1 = 0$

Let a point be $P (x, y)$ , such that

$S P = e P M$ , where $P M$ is perpendicular distance from $P (x, y)$ to directrix

$\Rightarrow \sqrt{(x - 1)^{2} + (y - 1)^{2}} = \sqrt{3} \times ∣ ∣ ∣ ∣ \frac{2 x + y - 3}{\sqrt{2^{2} + 1^{2}}} ∣ ∣ ∣ ∣$

$\Rightarrow \sqrt{(x - 1)^{2} + (y - 1)^{2}} = \frac{\sqrt{3}}{\sqrt{5}} | 2 x + y - 1 |$

$\Rightarrow (x - 1)^{2} + (y - 1)^{2} = \frac{3}{5} (2 x + y - 1)^{2}$
$\Rightarrow 5 (x^{2} - 2 x + 1 + y^{2} - 2 y + 1) = 3 (4 x^{2} + y^{2} + 1 + 4 x y - 2 y - 4 x)$
$\Rightarrow 5 x^{2} - 10 x + 5 y^{2} - 10 y + 10 = 12 x^{2} + 3 y^{2} + 3 + 12 x y - 6 y - 12 x$
$\Rightarrow 7 x^{2} - 2 y^{2} + 12 x y - 2 x + 4 y - 7 = 0$

$∴$ Equation of hyperbola is
$7 x^{2} + 12 x y - 2 y^{2} - 2 x + 4 y - 7 = 0$

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