<p>Let any double ordinate \(PNP'\) of the hyperbola \(\dfrac{x^2}{25} - \dfrac{y^2}{16} = 1\) be produced on both sides to meet the asymptotes in \(Q\) and \(Q'\), then \(\dfrac{PQ \cdot P'Q}{5}\) is equal to</p>
Step-by-Step Solution
Key Concept: A double ordinate is a chord perpendicular to the major axis. Use the asymptote equations y = ±(4/5)x and the hyperbola equation to find where the vertical line through P and P' meets the asymptotes, then calculate the required ratio.
<p><strong>Step 1:</strong> For hyperbola x²/25 - y²/16 = 1, we have a² = 25 (a = 5) and b² = 16 (b = 4).</p><p><strong>Step 2:</strong> Asymptotes are y = ±(4/5)x.</p><p><strong>Step 3:</strong> Let P(x₀, y₀) be a point on the hyperbola, so x₀²/25 - y₀²/16 = 1. Then P' = (x₀, -y₀) (double ordinate perpendicular to major axis).</p><p><strong>Step 4:</strong> The vertical line x = x₀ meets asymptote y = (4/5)x at Q(x₀, 4x₀/5) and asymptote y = -(4/5)x at Q'(x₀, -4x₀/5).</p><p><strong>Step 5:</strong> Calculate PQ = |y₀ - 4x₀/5| and P'Q' = |-y₀ - (-4x₀/5)| = |-y₀ + 4x₀/5|.</p><p><strong>Step 6:</strong> Since P is on hyperbola: x₀²/25 - y₀²/16 = 1, so 16x₀² - 25y₀² = 400.</p><p><strong>Step 7:</strong> PQ · P'Q = |4x₀/5 - y₀| · |4x₀/5 + y₀| = |(4x₀/5)² - y₀²| = |16x₀²/25 - y₀²| = |16x₀²/25 - (16x₀² - 400)/25| = |400/25| = 16.</p><p><strong>Step 8:</strong> Therefore, (PQ · P'Q)/5 = 16/5.</p><p>∴ Answer: <strong>16/5</strong></p>
Correct Answer: 16