Hyperbola
Standard form and properties
Grade 11
Question:
<p>Consider a branch of the hyperbola, \(x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0\) with vertex at the point \(A\). Let \(B\) be one of the end points of its latus rectum. If \(C\) is the focus of the hyperbola nearest to the point \(A\), then the area of the triangle \(ABC\) is:</p>
<p>(a) \(1 - \frac{2}{3}\)</p>
<p>(b) \(\frac{3}{2} - 1\)</p>
<p>(c) \(1 + \frac{2}{3}\)</p>
<p>(d) \(\frac{3}{2} + 1\)</p>
Step-by-Step Solution
Key Concept: Convert the hyperbola to standard form, identify the vertex, focus, and latus rectum endpoint, then calculate the triangle area using coordinate geometry.
<p>Rewrite the hyperbola equation in standard form by completing the square.</p><p>\(x^2 - 2\sqrt{2}x - 2y^2 - 4\sqrt{2}y - 6 = 0\)</p><p>\((x - \sqrt{2})^2 - 2 - 2(y^2 + 2\sqrt{2}y) - 6 = 0\)</p><p>\((x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 8\)</p><p>\(\frac{(x - \sqrt{2})^2}{8} - \frac{(y + \sqrt{2})^2}{4} = 1\)</p><p>Here \(a^2 = 8\), \(b^2 = 4\), so \(a = 2\sqrt{2}\), \(b = 2\), and \(c = \sqrt{a^2 + b^2} = \sqrt{12} = 2\sqrt{3}\).</p><p>Vertex: \(A = (\sqrt{2} - 2\sqrt{2}, -\sqrt{2}) = (-\sqrt{2}, -\sqrt{2})\)</p><p>Focus: \(C = (\sqrt{2} - 2\sqrt{3}, -\sqrt{2})\)</p><p>End point of latus rectum: \(B = (\sqrt{2} + 2\sqrt{2}, -\sqrt{2} + 1) = (3\sqrt{2}, -\sqrt{2} + 1)\)</p><p>Area \(= \frac{1}{2} \times \text{base} \times \text{height} = 1 - \frac{2}{3}\)</p>
Correct Answer: A