Mathematics

The key is recognizing that both curves hide their standard forms behind trigonometric coefficients—you'll need to extract $a$ and $b$ for each, then use the eccentricity formulas $e_{\text{hyperbola}} = \sqrt{1 + \frac{b^2}{a^2}}$ and $e_{\text{ellipse}} = \sqrt{1 - \frac{b^2}{a^2}}$. Once you have both eccentricities in terms of $\sec \alpha$, the given relationship $e_h = \sqrt{3} \cdot e_e$ becomes an equation you can solve for $\alpha$.

The key is recognizing that both curves hide their standard forms behind trigonometric coefficients—you'll need to extract $a$ and $b$ for each, then use the eccentricity formulas $e_{\text{hyperbola}} = \sqrt{1 + \frac{b^2}{a^2}}$ and $e_{\text{ellipse}} = \sqrt{1 - \frac{b^2}{a^2}}$. Once you have both eccentricities in terms of $\sec \alpha$, the given relationship $e_h = \sqrt{3} \cdot e_e$ becomes an equation you can solve for $\alpha$.