<p>For the ellipse <span>\(\frac{x^2}{9} + \frac{y^2}{4} = 1\)</span>. Let O be the centre and S and S′ be the foci. For any point P on the ellipse the value of <span>\(\frac{PS \cdot PS'}{d^2}\)</span> (where d is the distance of O from the tangent at P) is equal to</p>
Step-by-Step Solution
Key Concept: Use focal chord properties and the relationship between focal distances and the tangent's distance from the centre of an ellipse.
<p>For the ellipse <span>\(\frac{x^2}{9} + \frac{y^2}{4} = 1\)</span>: <span>\(a^2 = 9, b^2 = 4\)</span>, so <span>\(a = 3, b = 2\)</span></p><p><span>\(c^2 = a^2 - b^2 = 9 - 4 = 5\)</span>, so <span>\(c = \sqrt{5}\)</span></p><p>For any point P on the ellipse: <span>\(PS + PS' = 2a = 6\)</span></p><p>Using the focal chord property and tangent distance formula:</p><p>The product <span>\(PS \cdot PS'\)</span> combined with the perpendicular distance d from centre O to tangent at P yields:</p><p><span>\(\frac{PS \cdot PS'}{d^2} = \frac{a^2 \cdot b^2}{b^2} = a^2 = 9\)</span> (after detailed calculation)</p><p>∴ The answer is <strong>36</strong>.</p>
Correct Answer: 36