The length of latus rectum of the hyperbola $4 x^{2} - 9 y^{2} - 16 x - 54 y - 101 = 0$ is

\frac{8}{5}

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\frac{8}{7}

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\frac{8}{9}

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\frac{8}{3}

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Solution

The correct option is D $\frac{8}{3}$
$4 x^{2} - 9 y^{2} - 16 x - 54 y - 101 = 0$
$\Rightarrow 4 (x^{2} - 4 x) - 9 (y^{2} + 6 y) = 101$
$\Rightarrow 4 (x^{2} - 4 x + 4) - 9 (y^{2} + 6 y + 9) = 101 + 16 - 81 = 36$
$\Rightarrow \frac{(x - 2)^{2}}{3^{2}} - \frac{(y + 3)^{2}}{2^{2}} = 1$
$\Rightarrow a = 3, b = 2$
Thus length of latus rectum of the given hyperbola is $= \frac{2 b^{2}}{a} = \frac{8}{3}$

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