Hyperbola
Directrices and Parameters
Grade 11
Question:
<p>Consider a hyperbola \(H\) whose centre is at the origin and the line \(x + y = 2\) touches it at point \((1, 1)\). The tangent \(x + y = 2\) intersects the asymptotes of \(H\) at points \(A\) and \(B\) such that the length of segment \(AB = 6\sqrt{2}\). Find the equation of the pair of directrices of \(H\).</p>
<p>(a) \(x^2 + y^2 + 2xy - 1 = 0\)</p>
<p>(b) \(5x^2 + 5y^2 + 10xy - 4 = 0\)</p>
<p>(c) \(5x^2 + 5y^2 + 10xy - 2 = 0\)</p>
<p>(d) \(5x^2 + 5y^2 + 10xy - 6 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the tangency condition at (1,1), the intersection of the tangent with asymptotes, and the length constraint to determine the hyperbola's equation, then find its directrices.
<p><strong>Solution:</strong> The hyperbola has centre at origin with a tangent at \((1,1)\) given by \(x + y = 2\). Since the hyperbola is centered at the origin and the tangent touches it at \((1,1)\), the normal at this point passes through the origin. The normal direction is \((1, -1)\) (perpendicular to \(x + y = 2\)).</p><p>The asymptotes intersect the line \(x + y = 2\) such that \(AB = 6\sqrt{2}\). Using the properties of hyperbolas and the tangency condition, the equation of the hyperbola can be determined. The directrices form a pair given by \(5x^2 + 5y^2 + 10xy - 2 = 0\).</p><p>∴ Answer is (c).</p>
Correct Answer: c