The equation of the hyperbola with focus (1, | vedclass

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(A) Given,Focus $(S) = (1, 2)$,Eccentricity $(e) = \sqrt{3}$,and Directrix $2x + y - 1 = 0$. By the definition of a conic,$SP = e \cdot PM$,where $P(x

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$2y^2 - 12xy - 7x^2 + 2x - 14y + 22 = 0$

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(A) Given,Focus $(S) = (1, 2)$,Eccentricity $(e) = \sqrt{3}$,and Directrix $2x + y - 1 = 0$. By the definition of a conic,$SP = e \cdot PM$,where $P(x, y)$ is a point on the hyperbola. $SP^2 = e^2 \cdot PM^2$ $(x - 1)^2 + (y - 2)^2 = 3 \cdot \frac{(2x + y - 1)^2}{2^2 + 1^2}$ $x^2 - 2x + 1 + y^2 - 4y + 4 = \frac{3}{5}(4x^2 + y^2 + 1 + 4xy - 4x - 2y)$ $5(x^2 + y^2 - 2x - 4y + 5) = 3(4x^2 + y^2 + 4xy - 4x - 2y + 1)$ $5x^2 + 5y^2 - 10x - 20y + 25 = 12x^2 + 3y^2 + 12xy - 12x - 6y + 3$ Rearranging the terms: $12x^2 - 5x^2 + 3y^2 - 5y^2 + 12xy - 12x + 10x - 6y + 20y + 3 - 25 = 0$ $7x^2 - 2y^2 + 12xy - 2x + 14y - 22 = 0$ Multiplying by $-1$: $-7x^2 + 2y^2 - 12xy + 2x - 14y + 22 = 0$ Thus,the equation is $2y^2 - 12xy - 7x^2 + 2x - 14y + 22 = 0$.

The equation of the hyperbola with focus $(1, 2)$,eccentricity $e = \sqrt{3}$,and directrix $2x + y = 1$ is given by:

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