Ellipse
Tangent to Ellipse
Grade 11
Question:
<p>If the line \(2px + y\sqrt{1-p^2} = 1\) always touches the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) \(\forall\, p \in (-1,1) - \{0\}\). The eccentricity of this ellipse, is</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\dfrac{\sqrt{7}}{3}\)</p>
<p>\(\dfrac{\sqrt{7}}{4}\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
Step-by-Step Solution
Key Concept: A line touches an ellipse if the discriminant of their intersection equals zero. Since the line 2px + y√(1-p²) = 1 touches the ellipse for all p ∈ (-1,1)-{0}, we use the tangency condition: the line must satisfy the tangent equation a²(2p)² + b²(√(1-p²))² = 1 for all valid p.
<p><strong>Step 1:</strong> For the line 2px + y√(1-p²) = 1 to be tangent to the ellipse x²/a² + y²/b² = 1, use the tangency condition: c₁²a² + c₂²b² = 1, where c₁ and c₂ are coefficients of x and y.</p><p>Here: (2p)²a² + (√(1-p²))²b² = 1</p><p>Expanding: 4a²p² + b²(1-p²) = 1</p><p><strong>Step 2:</strong> Rearrange: 4a²p² + b² - b²p² = 1</p><p>⟹ p²(4a² - b²) + b² = 1</p><p><strong>Step 3:</strong> Since this must hold for ALL p ∈ (-1,1)-{0}, the coefficient of p² and the constant term must independently satisfy the equation:</p><p>• Coefficient of p²: 4a² - b² = 0 ⟹ b² = 4a²</p><p>• Constant term: b² = 1 ⟹ b = 1</p><p><strong>Step 4:</strong> From b² = 4a²: 1 = 4a² ⟹ a² = 1/4</p><p>Thus: a = 1/2, b = 1</p><p><strong>Step 5:</strong> Find eccentricity: e² = 1 - a²/b² = 1 - (1/4)/1 = 1 - 1/4 = 3/4</p><p>∴ e = √(3/4) = √3/2</p><p><strong>Answer: D (√3/2)</strong></p>
Correct Answer: D