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 :

The correct option is D $6$
$T:\frac{x{x}_1}{4}-\frac{y{y}_1}{2}=1....\left(1\right)$

$t:2x-y=0$ is parallel to $T$

$\Rightarrow T:2x-y=\lambda ....\left(2\right)$

Now comparing the slopes
$x_1=4y_1$
$(x_1,y_1)\text{lies on hyperbola}\Rightarrow 4y_1^2-\frac{y_1^2}{2}=1\Rightarrow 7y_1^2=2$
Now
$x_1^2+5y_1^2=21y_1^2=6$