Question:

<p><strong>Step 1: Identify hyperbola parameters</strong></p><p>From $\frac{x^2}{9} - \frac{y^2}{3} = 1$: we have $a^2 = 9$ and $b^2 = 3$, so $a = 3$ and $b = \sqrt{3}$</p><p><strong>Step 2: Check statement (a) - Acute angle between asymptotes</strong></p><p>Asymptotes: $y = \pm\frac{b}{a}x = \pm\frac{\sqrt{3}}{3}x = \pm\frac{1}{\sqrt{3}}x$</p><p>For acute angle θ between asymptotes: $\tan(\theta/2) = \frac{b}{a} = \frac{\sqrt{3}}{3}$</p><p>So $\theta/2 = 30°$, giving $\theta = 60°$ ✓ (Statement (a) is CORRECT)</p><p><strong>Step 3: Check statement (b) - Eccentricity</strong></p><p>$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{9}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$</p><p>Given statement: $e = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$ ✗ (Statement (b) is INCORRECT - this does not match)</p><p><strong>Step 4: Verify statement (c) - Length of latus rectum</strong></p><p>Latus rectum $= \frac{2b^2}{a} = \frac{2 \times 3}{3} = 2$ ✓ (Statement (c) is CORRECT)</p><p><strong>Step 5: Verify statement (d) - Product of perpendicular distances</strong></p><p>For any point P on the hyperbola, the product of perpendicular distances to the asymptotes $= \frac{a^2b^2}{a^2+b^2} = \frac{9 \times 3}{9+3} = \frac{27}{12} = \frac{9}{4} = 2.25$</p><p>Length of latus rectum = 2, and $2.25 > 2$, so the product is GREATER than latus rectum length ✓ (Statement (d) is CORRECT)</p><p><strong>∴ Answer:</strong> b</p>