# 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 $\left ( x_1,y_1 \right ).$ Then $x_{1}^{2}+5y_{1}^{2}$ is equal to:   Option: 1 6 Option: 2 8 Option: 3 10 Option: 4 5

Slope of tangent is 2, Tangent of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left ( x_1,y_1 \right )$ is

$\\\frac{x x_{1}}{4}-\frac{y y_{1}}{2}=1 \quad(T=0) \\ \text { Slope: } \frac{1}{2} \frac{x_{1}}{y_{1}}=2 \Rightarrow x_{1}=4 y_{1} \\$

$\left ( x_1,y_1 \right )$ lies on the parabola

$\Rightarrow \frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1$

From above equations

$\\\frac{\left(4 y_{1}\right)^{2}}{4}-\frac{y_{1}^{2}}{2}=1 \Rightarrow 4 y_{1}^{2}-\frac{y_{1}^{2}}{2}=1 \\ \Rightarrow 7 y_{1}^{2}=2 \Rightarrow y_{1}^{2}=\frac{2}{7} \\ \text { Now } x_{1}^{2}+5 y_{1}^{2}=\left(4 y_{1}\right)^{2}+5 y_{1}^{2} \\ =(21) y_{1}^{2}=21 \times \frac{2}{7}=6$

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