The area (in sq. units) of the triangle formed by the lines y = 2x,...

Equation of the tangent at point $\left(\sqrt{5}, 4\right)$ on $4x^{2} - y^{2} = 4$ is $4x\sqrt{5} - 4y = 4 ⇒ \sqrt{5}x - y = 1$.
Now, point of intersection of y = \sqrt{5}x - 1 & y = 2x is \left(\dfrac{1}{\sqrt{5} - 2}, \dfrac{2}{\sqrt{5} - 2}\right)$.<br />Point of intersection of$y = \sqrt{5}x - 1$and$y = - 2xis& \left(\dfrac{1}{\sqrt{5} + 2}, \dfrac{- 2}{\sqrt{5} + 2}\right) .
Point of intersection of $y = 2x$ & $y = - 2x$ is \left(0,0\right)$<br />Now, area of the triangle formed by the points \left(0,0\right), \left(\dfrac{1}{\sqrt{5} - 2}, \dfrac{2}{\sqrt{5} - 2}\right), \left(\dfrac{1}{\sqrt{5} + 2}, \dfrac{- 2}{\sqrt{5} + 2}\right)<br />$= \left|\dfrac{1}{2}\left|$\begin{cases}0 & 0 & 1 \\ 1\left(\sqrt{5} -2\right) & 2\left(\sqrt{5} -2\right) & 1 \\ 1\left(\sqrt{5} +2\right) & - 2\left(\sqrt{5} +2\right) & 1\end{cases}$\right|\right| = \left|\dfrac{1}{2} \times1 \times \left(-2 - 2\right)\right| = 2& sq.units