Ellipse
Tangent to Ellipse and Eccentricity
Grade 11
Question:
<p>A line <span>\( y = \dfrac{-2p}{\sqrt{1-p^2}}x + \dfrac{1}{\sqrt{1-p^2}} \)</span> is tangent to an ellipse <span>\( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \)</span> for all <span>\( p \in (-1,1) \setminus \{0\} \)</span>. Find the eccentricity of the ellipse.</p>
<p>\( \dfrac{1}{2} \)</p>
<p>\( \dfrac{\sqrt{3}}{2} \)</p>
<p>\( \dfrac{1}{\sqrt{2}} \)</p>
<p>\( \dfrac{2}{\sqrt{3}} \)</p>
Step-by-Step Solution
Key Concept: For a line y = mx + c to be tangent to ellipse x²/a² + y²/b² = 1 for all values of parameter p, the tangency condition c² = a²m² + b² must hold identically. Extract m and c from the given line, apply this condition, and use the relationship between a, b, and eccentricity.
<p><strong>Step 1:</strong> Identify m and c from the given line y = -2p/√(1-p²) · x + 1/√(1-p²)</p><p>Here: m = -2p/√(1-p²) and c = 1/√(1-p²)</p><p><strong>Step 2:</strong> Apply tangency condition c² = a²m² + b²</p><p>1/(1-p²) = a² · 4p²/(1-p²) + b²</p><p><strong>Step 3:</strong> Simplify by multiplying both sides by (1-p²)</p><p>1 = 4a²p² + b²(1-p²)</p><p>1 = 4a²p² + b² - b²p²</p><p>1 = b² + (4a² - b²)p²</p><p><strong>Step 4:</strong> For this to hold for ALL p ∈ (-1,1)\{0}, both the constant and coefficient of p² must match:</p><p>Constant term: b² = 1, so b = 1</p><p>Coefficient of p²: 4a² - b² = 0, so 4a² = 1, thus a² = 1/4</p><p><strong>Step 5:</strong> Calculate eccentricity using e² = 1 - b²/a²</p><p>e² = 1 - 1/(1/4) = 1 - 4 = -3 (impossible)</p><p>Correct application: e² = 1 - a²/b² = 1 - (1/4)/1 = 3/4</p><p>∴ e = √(3/4) = <strong>√3/2</strong></p><p><strong>Answer: B</strong></p>
Correct Answer: B