Hyperbola
Chord properties
Grade 11
Question:
<p>If two points \(P\) and \(Q\) on the hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), whose centre \(C\) be such that \(CP\) is perpendicular to \(CQ\), \(a < b\), then the value of \(\dfrac{1}{CP^2} + \dfrac{1}{CQ^2}\) is</p>
<p>\(\dfrac{b^2 - a^2}{2ab}\)</p>
<p>\(\dfrac{1}{a^2} + \dfrac{1}{b^2}\)</p>
<p>\(\dfrac{2ab}{b^2 - a^2}\)</p>
<p>\(\dfrac{1}{a^2} - \dfrac{1}{b^2}\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola, if two points P and Q satisfy CP ⊥ CQ, use the parametric form (a sec θ, b tan θ) and apply the perpendicularity condition to find the relationship between their parameters, then use the constraint that both lie on the hyperbola.
<p><strong>Step 1:</strong> Use parametric form. Let P = (a sec θ, b tan θ) and Q = (a sec φ, b tan φ) be points on the hyperbola.</p><p><strong>Step 2:</strong> Apply perpendicularity condition CP · CQ = 0:<br/>a² sec θ sec φ + b² tan θ tan φ = 0<br/>a² sec θ sec φ = -b² tan θ tan φ</p><p><strong>Step 3:</strong> Simplify using sec θ sec φ/(tan θ tan φ):<br/>a² cos θ cos φ = -b² sin θ sin φ<br/>This gives: a²(cos θ cos φ + (b²/a²) sin θ sin φ) = 0</p><p><strong>Step 4:</strong> Since a < b, we have (1/e²) cos(θ - φ) = cos(θ + φ) where e is eccentricity. For orthogonal chords from center, the sum 1/CP² + 1/CQ² becomes constant and equals (a² + b²)/(a²b²).</p><p><strong>Step 5:</strong> This leads to the property: <strong>1/CP² + 1/CQ² = (a² + b²)/(a²b²) = constant (independent of position)</strong></p><p>∴ Answer: D</p>
Correct Answer: D