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 for all values of a parameter if and only if the line satisfies the tangency condition for every parameter value. Rewrite the given line equation in the form of a family of tangent lines to identify the constraint on a and b.
<p><strong>Step 1:</strong> Recognize that the line <strong>2px + y√(1-p²) = 1</strong> must be tangent to the ellipse for all p ∈ (-1,1) - {0}.</p><p><strong>Step 2:</strong> Rewrite using substitution: Let p = cos θ. Then √(1-p²) = sin θ, so the line becomes:</p><p><strong>2x cos θ + y sin θ = 1</strong></p><p><strong>Step 3:</strong> Recognize this as a parametric family. For a line lx + my = 1 to always touch the ellipse x²/a² + y²/b² = 1, the tangency condition is:</p><p><strong>a²l² + b²m² = 1</strong></p><p><strong>Step 4:</strong> Here, l = 2cos θ and m = sin θ. Substituting into the tangency condition:</p><p><strong>a²(2cos θ)² + b²(sin θ)² = 1</strong></p><p><strong>4a² cos² θ + b² sin² θ = 1</strong></p><p><strong>Step 5:</strong> For this to hold for ALL θ (or all p), comparing coefficients:</p><p><strong>4a² = 1 and b² = 1</strong></p><p><strong>Therefore: a² = 1/4, so a = 1/2, and b = 1</strong></p><p><strong>Step 6:</strong> Since b > a, we have b² = a² + c², so:</p><p><strong>1 = 1/4 + c²</strong></p><p><strong>c² = 3/4, c = √3/2</strong></p><p><strong>Step 7:</strong> Eccentricity e = c/b = (√3/2)/1 = <strong>√3/2</strong></p><p>∴ Answer: D</p>
Correct Answer: D