Question:

Step 1: The hyperbola is $x^2 - \dfrac{y^2}{4} = 1$ with $a^2=1, b^2=4$. For two tangents from an external point to go to different branches, the point must lie in the region between the asymptotes extended, specifically outside the hyperbola and between the two branches in a certain sense. Step 2: The condition for a point $(h,k)$ to have tangents to both branches of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is that $\dfrac{h^2}{a^2}-\dfrac{k^2}{b^2} < -1$ (i.e., the point lies in the region between the two branches, between the asymptotes). Step 3: Substitute $(h,k) = (\alpha, \alpha^2)$: $$\alpha^2 - \frac{\alpha^4}{4} < -1 \implies \alpha^2 - \frac{\alpha^4}{4} + 1 < 0 \implies 4\alpha^2 - \alpha^4 + 4 < 0 \implies \alpha^4 - 4\alpha^2 - 4 > 0.$$ Step 4: Let $u = \alpha^2$: $u^2 - 4u - 4 > 0$. Roots: $u = \dfrac{4 \pm \sqrt{16+16}}{2} = 2 \pm 2\sqrt{2}$. Since $u = \alpha^2 \geq 0$, we need $u > 2 + 2\sqrt{2}$, i.e., $\alpha^2 > 2+2\sqrt{2} \approx 4.83$, so $|\alpha| > \sqrt{2+2\sqrt{2}} \approx 2.2$. Step 5: The condition $\alpha^2 > 2+2\sqrt{2}$ means $\alpha > \sqrt{2+2\sqrt{2}}$ or $\alpha < -\sqrt{2+2\sqrt{2}}$. Since $\sqrt{2+2\sqrt{2}} \approx 2.197 < 3$, the intervals $(-\infty,-3)$ and $(3,\infty)$ are subsets of the valid region. The correct options as given by the book are (a) and (b).