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A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left(x_{1}, y_{1}\right) .$ Then $x_{1}^{2}+5 y_{1}^{2}$ is equal to
$5$
$6$
$8$
$10$
Solution
Slope of tangent is $2,$ Tangent of hyperbola
$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left(x_{1}, y_{1}\right)$ is
$\frac{\mathrm{xx}_{1}}{4}-\frac{\mathrm{yy}_{1}}{2}=1 \quad(\mathrm{T}=0)$
Slope $: \frac{1}{2} \frac{x_{1}}{y_{1}}=2 \Rightarrow\left[x_{1}=4 y_{1}\right.$
$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ lies on hyperbola
$\Rightarrow\left[\frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1\right.$
From (1)$\&(2)$
$\frac{\left(4 y_{1}\right)^{2}}{4}-\frac{y_{1}^{2}}{2}=1 \Rightarrow 4 y_{1}^{2}-\frac{y_{1}^{2}}{2}=1$
$\Rightarrow 7 y_{1}^{2}=2 \Rightarrow y_{1}^{2}=2 / 7$
Now $x_{1}^{2}+5 y_{1}^{2}=\left(4 y_{1}\right)^{2}+5 y_{1}^{2}$
$=(21) \mathrm{y}_{1}^{2}=21 \times \frac{2}{7}=6$
