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 :
Correct Option: 1
The tangent to the hyperbola at the point $\left(x_{1}, y_{1}\right)$ is,
$x x_{1}-2 y y_{1}-4=0$
The given equation of tangent is
$2 x-y=0$
$\Rightarrow \frac{x_{1}}{2 y_{1}}=2$
$\Rightarrow x_{1}=4 y_{1}$ ...(i)
Since, point $\left(x_{1}, y_{1}\right)$ lie on hyperbola.
$\therefore \frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}-1=0$ ...(ii)
On solving eqs. (i) and (ii)
$y_{1}^{2}=\frac{2}{7}, x_{1}^{2}=\frac{32}{7}$
$\therefore \quad x_{1}^{2}+5 y_{1}^{2}=\frac{32}{7}+5 \times \frac{2}{7}=6$
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