The eccentricity of the hyperbola 5x^2 - 4y^2 | vedclass

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(B) Given equation of the hyperbola is $5x^2 - 4y^2 + 20x + 8y = 4$.Rearranging the terms,we get $5(x^2 + 4x) - 4(y^2 - 2y) = 4$.Completing the square

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$\frac{3}{2}$

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(B) Given equation of the hyperbola is $5x^2 - 4y^2 + 20x + 8y = 4$.Rearranging the terms,we get $5(x^2 + 4x) - 4(y^2 - 2y) = 4$.Completing the square,we have $5(x^2 + 4x + 4) - 4(y^2 - 2y + 1) = 4 + 20 - 4$.This simplifies to $5(x + 2)^2 - 4(y - 1)^2 = 20$.Dividing by $20$,we get $\frac{(x + 2)^2}{4} - \frac{(y - 1)^2}{5} = 1$.Comparing this with the standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = 5$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.$e = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

The eccentricity of the hyperbola $5x^2 - 4y^2 + 20x + 8y = 4$ is

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