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(C) The eccentricity $e'$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ is given by:$e' = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \frac

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$6$

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(C) The eccentricity $e'$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ is given by:$e' = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$Given that the eccentricity $e$ of the hyperbola $H$ is the reciprocal of $e'$,we have:$e = \frac{1}{e'} = \sqrt{2}$For a hyperbola with transverse axis along the $X$-axis,$e = \sqrt{1 + \frac{b^2}{a^2}}$. Thus:$\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2} \implies 1 + \frac{b^2}{a^2} = 2 \implies \frac{b^2}{a^2} = 1 \implies b^2 = a^2$The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$.Since the point $(5, 4)$ lies on $H$:$\frac{25}{a^2} - \frac{16}{a^2} = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9 \implies a = 3$The length of the transverse axis is $2a = 2 	imes 3 = 6$.

Let $X$-axis be the transverse axis and $Y$-axis be the conjugate axis of a hyperbola $H$. Let the eccentricity of $H$ be the reciprocal of the eccentricity of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$. If $(5, 4)$ is a point on $H$,then the length of the transverse axis of $H$ is

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