The eccentricity of the hyperbola $4 x^{2} - y^{2} - 8 x - 8 y - 28$ is equal to ___ .

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$\sqrt{5}$

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$\sqrt{7}$

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Solution

The correct option is B
$\sqrt{5}$

$4 x^{2} - y^{2} - 8 x - 8 y - 28 {4 (x - 1)^{2} - 4} - {(y + 4)^{2} - 16} - 28 = 0 4 (x - 1)^{2} - (y + 4)^{2} = 16 \frac{(x - 1)^{2}}{4} - \frac{(y + 4)^{2}}{16} = 1$

For the above hyperbola, the eccentricity is given by
$e = \frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{4 + 16}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$

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Similar questions

4 x^{2} - y^{2} - 8 x - 8 y - 28 = 0

4 x^{2} - 9 y^{2} - 8 x = 32

5 x^{2} - 4 y^{2} + 20 x + 8 y = 4

Then eccentricity of the Ellipse $4 x^{2} + y^{2} - 8 x - 2 y + 1$ = 0

\sqrt{2} .

\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1

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