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(B) The slope of the line $2x - y = 0$ is $2$. Since the tangent is parallel to this line,its slope is $m = 2$.The equation of the tangent to the hype

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$6$

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(B) The slope of the line $2x - y = 0$ is $2$. Since the tangent is parallel to this line,its slope is $m = 2$.The equation of the tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$ is given by $\frac{xx_{1}}{4} - \frac{yy_{1}}{2} = 1$.The slope of this tangent is $\frac{x_{1}/4}{y_{1}/2} = \frac{x_{1}}{2y_{1}}$.Equating the slopes: $\frac{x_{1}}{2y_{1}} = 2 \Rightarrow x_{1} = 4y_{1} \quad (1)$.Since $(x_{1}, y_{1})$ lies on the hyperbola,we have $\frac{x_{1}^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \quad (2)$.Substituting $(1)$ into $(2)$: $\frac{(4y_{1})^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \Rightarrow 4y_{1}^{2} - \frac{y_{1}^{2}}{2} = 1$.$\frac{7y_{1}^{2}}{2} = 1 \Rightarrow y_{1}^{2} = \frac{2}{7}$.Now,calculate $x_{1}^{2} + 5y_{1}^{2} = (4y_{1})^{2} + 5y_{1}^{2} = 16y_{1}^{2} + 5y_{1}^{2} = 21y_{1}^{2}$.Substituting $y_{1}^{2} = \frac{2}{7}$: $21 	imes \frac{2}{7} = 3 	imes 2 = 6$.

$A$ line parallel to the straight line $2x - y = 0$ is tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$. Then $x_{1}^{2} + 5y_{1}^{2}$ is equal to:

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