The eccentricity of the hyperbola 4x2−y2−8x−8y−28

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Question

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

Options:

A . 2

B .

\sqrt{5}

C . 2

D .

\sqrt{7}

Answer: Option B
:
B

$4 x^{2} - y^{2} - 8 x - 8 y - 28 14 (x - 1)^{2} - 4} - 1 (y + 4)^{2} - 16} - 28 = 0 4 (x - 1)^{2} - (y + 4)^{2} = 16 \frac{(x - 1)^{2}}{4} - \frac{(y + 4)^{2}}{16} = 1$

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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