Question:

<p>For the hyperbola \(\frac{x^2}{9} - \frac{y^2}{3} = 1\), we have \(a^2 = 9\), \(b^2 = 3\), so \(a = 3\), \(b = \sqrt{3}\). (a) The asymptotes are \(y = \pm\frac{b}{a}x = \pm\frac{\sqrt{3}}{3}x\). The acute angle \(\alpha\) between asymptotes satisfies \(\tan\alpha = \frac{2ab}{a^2-b^2} = \frac{2 \cdot 3 \cdot \sqrt{3}}{9-3} = \frac{6\sqrt{3}}{6} = \sqrt{3}\), so \(\alpha = 60°\). This is correct. (b) Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\), not \(\frac{4}{\sqrt{3}}\). This is incorrect. (c) Latus rectum \(= \frac{2b^2}{a} = \frac{2 \cdot 3}{3} = 2\). This is correct. (d) The product of perpendicular distances from a point on the hyperbola to its asymptotes is \(\frac{a^2b^2}{a^2+b^2} = \frac{9 \cdot 3}{12} = \frac{9}{4} = 2.25\), which is greater than 2. So this statement needs verification but option (b) is clearly incorrect.</p>