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(D) Let the equation of the tangent passing through $(0, 1)$ with slope $m$ be $y - 1 = m(x - 0)$,which simplifies to $y = mx + 1$.Substituting this i

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$9x - 2y + 2 = 0$

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(D) Let the equation of the tangent passing through $(0, 1)$ with slope $m$ be $y - 1 = m(x - 0)$,which simplifies to $y = mx + 1$.Substituting this into the hyperbola equation $45x^2 - 4y^2 = 5$:$45x^2 - 4(mx + 1)^2 = 5$$45x^2 - 4(m^2x^2 + 2mx + 1) = 5$$(45 - 4m^2)x^2 - 8mx - 9 = 0$For the line to be a tangent,the discriminant $D$ must be $0$:$D = (-8m)^2 - 4(45 - 4m^2)(-9) = 0$$64m^2 + 36(45 - 4m^2) = 0$$64m^2 + 1620 - 144m^2 = 0$$-80m^2 + 1620 = 0$$80m^2 = 1620$$m^2 = \frac{1620}{80} = \frac{81}{4}$$m = \pm \frac{9}{2}$Substituting $m = \frac{9}{2}$ into $y = mx + 1$:$y = \frac{9}{2}x + 1$ $\Rightarrow 2y = 9x + 2$ $\Rightarrow 9x - 2y + 2 = 0$Substituting $m = -\frac{9}{2}$ into $y = mx + 1$:$y = -\frac{9}{2}x + 1$ $\Rightarrow 2y = -9x + 2$ $\Rightarrow 9x + 2y - 2 = 0$Thus,the correct option is $D$.

The equation of one of the tangents drawn from the point $(0, 1)$ to the hyperbola $45x^2 - 4y^2 = 5$ is

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