Question:

<p><strong>Step 1: Write the hyperbola in standard form.</strong></p><p>Given: $x^2 - 3y^2 = 3$</p><p>Dividing by 3: $\frac{x^2}{3} - \frac{y^2}{1} = 1$</p><p>So $a^2 = 3$ and $b^2 = 1$, giving $a = \sqrt{3}$ and $b = 1$.</p><p><strong>Step 2: Find the equation of asymptotes.</strong></p><p>The asymptotes of $\frac{x^2}{3} - \frac{y^2}{1} = 1$ are: $y = \pm\frac{b}{a}x = \pm\frac{1}{\sqrt{3}}x$</p><p>Or: $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$</p><p><strong>Step 3: Find the tangent to the hyperbola at $(\sqrt{3}, 0)$.</strong></p><p>Using the tangent formula for $\frac{x^2}{3} - \frac{y^2}{1} = 1$ at point $(x_0, y_0)$:</p><p>$\frac{xx_0}{3} - \frac{yy_0}{1} = 1$</p><p>At $(\sqrt{3}, 0)$: $\frac{x \cdot \sqrt{3}}{3} - \frac{y \cdot 0}{1} = 1$</p><p>$\frac{x}{\sqrt{3}} = 1 \Rightarrow x = \sqrt{3}$</p><p>The tangent is the vertical line: $x = \sqrt{3}$</p><p><strong>Step 4: Find intersection points of tangent with asymptotes.</strong></p><p>Intersection with $x - \sqrt{3}y = 0$: $\sqrt{3} - \sqrt{3}y = 0 \Rightarrow y = 1$. Point: $(\sqrt{3}, 1)$</p><p>Intersection with $x + \sqrt{3}y = 0$: $\sqrt{3} + \sqrt{3}y = 0 \Rightarrow y = -1$. Point: $(\sqrt{3}, -1)$</p><p>Intersection of asymptotes: $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$ gives $x = 0, y = 0$. Point: $(0, 0)$</p><p><strong>Step 5: Calculate side lengths of triangle with vertices $(0,0)$, $(\sqrt{3}, 1)$, $(\sqrt{3}, -1)$.</strong></p><p>Side 1: From $(0,0)$ to $(\sqrt{3}, 1)$: $\sqrt{3 + 1} = 2$</p><p>Side 2: From $(0,0)$ to $(\sqrt{3}, -1)$: $\sqrt{3 + 1} = 2$</p><p>Side 3: From $(\sqrt{3}, 1)$ to $(\sqrt{3}, -1)$: $\sqrt{0 + 4} = 2$</p><p><strong>Step 6: Verify triangle properties.</strong></p><p>All three sides equal 2, so the triangle is equilateral.</p><p>$∴$ Answer: b</p>