The eccentricity of the conjugate hyperbola o | vedclass

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(B) Given equation: $16x^2 - 9y^2 - 32x - 36y - 164 = 0$Rearranging terms: $16(x^2 - 2x) - 9(y^2 + 4y) = 164$Completing the square: $16(x^2 - 2x + 1)

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$\frac{5}{4}$

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(B) Given equation: $16x^2 - 9y^2 - 32x - 36y - 164 = 0$Rearranging terms: $16(x^2 - 2x) - 9(y^2 + 4y) = 164$Completing the square: $16(x^2 - 2x + 1) - 9(y^2 + 4y + 4) = 164 + 16 - 36$$16(x - 1)^2 - 9(y + 2)^2 = 144$Dividing by $144$: $\frac{(x - 1)^2}{9} - \frac{(y + 2)^2}{16} = 1$Here,$a^2 = 9$ and $b^2 = 16$. The eccentricity $e$ of the hyperbola is $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.The conjugate hyperbola is $\frac{(y + 2)^2}{16} - \frac{(x - 1)^2}{9} = 1$.For the conjugate hyperbola,the roles of $a$ and $b$ are swapped,so its eccentricity $e'$ is $\sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

The eccentricity of the conjugate hyperbola of $16x^2 - 9y^2 - 32x - 36y - 164 = 0$ is:

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