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(A) The equation of a hyperbola with asymptotes parallel to $2x + 3y = 0$ and $3x + 2y = 0$ is given by $(2x + 3y + c_1)(3x + 2y + c_2) = k$. Since th

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$(2x + 3y - 8)(3x + 2y - 7) = 154$

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(A) The equation of a hyperbola with asymptotes parallel to $2x + 3y = 0$ and $3x + 2y = 0$ is given by $(2x + 3y + c_1)(3x + 2y + c_2) = k$. Since the center is $(1, 2)$,the lines must pass through $(1, 2)$. For the first line: $2(1) + 3(2) + c_1 = 0 \implies 2 + 6 + c_1 = 0 \implies c_1 = -8$. For the second line: $3(1) + 2(2) + c_2 = 0 \implies 3 + 4 + c_2 = 0 \implies c_2 = -7$. So the equation is $(2x + 3y - 8)(3x + 2y - 7) = k$. Since it passes through $(5, 3)$,substitute $x = 5, y = 3$: $(2(5) + 3(3) - 8)(3(5) + 2(3) - 7) = k$ $(10 + 9 - 8)(15 + 6 - 7) = k$ $(11)(14) = k \implies k = 154$. Thus,the equation is $(2x + 3y - 8)(3x + 2y - 7) = 154$.

The asymptotes of a hyperbola are parallel to $2x + 3y = 0$ and $3x + 2y = 0$. The equation of the hyperbola whose center is at $(1, 2)$ and which passes through $(5, 3)$ is:

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