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Q.
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 :
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
$\frac{ xx _{1}}{4}-\frac{ yy _{1}}{2}=1 ( T =0)$
Slope : $\frac{1}{2} \frac{ x _{1}}{ y _{1}}=2 $
$\Rightarrow x _{1}=4 y _{1} \dots$(1)
$\left( x _{1}, y _{1}\right)$ lies on hyperbola
$\Rightarrow \frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1 \dots$(2)
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 $x_{1}^{2}+5 y_{1}^{2}=\left(4 y_{1}\right)^{2}+5 y_{1}^{2}$
$=(21) y _{1}^{2}=21 \times \frac{2}{7}=6$