Question:

<p>The hyperbola \(x^2 - 3y^2 = 3\) can be written as \(\frac{x^2}{3} - \frac{y^2}{1} = 1\), so \(a^2 = 3\), \(b^2 = 1\). The asymptotes are \(y = \pm\frac{1}{\sqrt{3}}x\). At the point \((\sqrt{3}, 0)\), differentiating implicitly: \(2x - 6yy' = 0\) gives \(y' = \frac{x}{3y}\). At \((\sqrt{3}, 0)\), the derivative is undefined (vertical tangent), but we can write the tangent as \(x = \sqrt{3}\). The asymptotes are \(y = \frac{x}{\sqrt{3}}\) and \(y = -\frac{x}{\sqrt{3}}\). Intersection of \(x = \sqrt{3}\) with \(y = \frac{x}{\sqrt{3}}\) gives \((\sqrt{3}, 1)\). Intersection of \(x = \sqrt{3}\) with \(y = -\frac{x}{\sqrt{3}}\) gives \((\sqrt{3}, -1)\). The three vertices are \((\sqrt{3}, 0)\), \((\sqrt{3}, 1)\), \((\sqrt{3}, -1)\). This forms a right isosceles triangle with the right angle at \((\sqrt{3}, 0)\) and two equal legs of length 1.</p>