The equation of the pair of asymptotes of the | vedclass

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(D) The given equation of the hyperbola is $4x^2 - 9y^2 - 24x - 36y - 36 = 0$. Rearranging the terms,we get $4(x^2 - 6x) - 9(y^2 + 4y) = 36$. Completi

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$4x^2 - 9y^2 - 24x - 36y = 0$

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(D) The given equation of the hyperbola is $4x^2 - 9y^2 - 24x - 36y - 36 = 0$. Rearranging the terms,we get $4(x^2 - 6x) - 9(y^2 + 4y) = 36$. Completing the square,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36$. This simplifies to $4(x - 3)^2 - 9(y + 2)^2 = 36$. Dividing by $36$,we get $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1$. The equation of the pair of asymptotes is obtained by setting the constant term to zero: $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 0$. Multiplying by $36$,we get $4(x - 3)^2 - 9(y + 2)^2 = 0$. Expanding this,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 0$. $4x^2 - 24x + 36 - 9y^2 - 36y - 36 = 0$. Thus,the equation is $4x^2 - 9y^2 - 24x - 36y = 0$.

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is

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