<p>If the line \(x - 2y = 12\) is tangent to the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) at the point \(\left(3, \dfrac{-9}{2}\right)\), then the length of the latus rectum of the ellipse is</p>
Step-by-Step Solution
Key Concept: For a tangent line to an ellipse at point (x₀, y₀), use the tangent equation xx₀/a² + yy₀/b² = 1 and match coefficients with the given line equation to find a² and b².
<p><strong>Step 1:</strong> Verify the point lies on the line: 3 - 2(-9/2) = 3 + 9 = 12 ✓</p><p><strong>Step 2:</strong> The tangent to ellipse x²/a² + y²/b² = 1 at point (x₀, y₀) is: (xx₀)/a² + (yy₀)/b² = 1</p><p>At point (3, -9/2): (3x)/a² + (-9y/2)/b² = 1</p><p><strong>Step 3:</strong> Rewrite the given line x - 2y = 12 in the form matching the tangent:</p><p>Dividing by 12: x/12 - y/6 = 1, or (3x)/36 + (-9y/2)/27 = 1</p><p><strong>Step 4:</strong> Compare coefficients with (3x)/a² + (-9y/2)/b² = 1:</p><p>• 3/a² = 3/36 ⟹ a² = 36</p><p>• (-9/2)/b² = (-9/2)/27 ⟹ b² = 27</p><p><strong>Step 5:</strong> Length of latus rectum = 2b²/a = 2(27)/6 = 9</p><p>∴ Answer: <strong>9</strong></p>
Correct Answer: 9