If the eccentricity of the standard hyperbola passing

Let equation of hyperbola be

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ....(i)

$\because e=\sqrt{1+\frac{b^{2}}{a^{2}}} \Rightarrow b^{2}=a^{2}\left(e^{2}-1\right)$

$e=2 \Rightarrow b^{2}=3 a^{2}$ ...(ii)

Equation (i) passes through $(4,6)$,

$\therefore \frac{16}{a^{2}}-\frac{36}{b^{2}}=1$ ...(iii)

On solving (i) and (ii), we get

$a^{2}=4, b^{2}=12$

Now equation of hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$

Now equation of tangent to the hyperbola at $(4,6)$ is

$\frac{4 x}{4}-\frac{6 y}{12}=1 \Rightarrow x-\frac{y}{2}=1 \Rightarrow 2 x-y=2$