Ellipse
Eccentricity relations between conics
Grade 11
Question:
<p>An ellipse and a hyperbola have the same foci. If <em>e</em><sub>1</sub> is the eccentricity of the ellipse and <em>e</em><sub>2</sub> is the eccentricity of the hyperbola, and <em>D</em> = <em>e</em><sub>2</sub> − <em>e</em><sub>1</sub>, then for \(\frac{1}{2} < e_1 < 1\), the range of <em>D</em> is</p>
<p>\(\left(\frac{1}{2}, \infty\right)\)</p>
<p>\(\left(\frac{1}{4}, \infty\right)\)</p>
<p>\((0, \infty)\)</p>
<p>\((1, \infty)\)</p>
Step-by-Step Solution
Key Concept: For an ellipse and hyperbola sharing the same foci, their focal distances are equal (c is constant). Use e₁ = c/a₁ for ellipse and e₂ = c/a₂ for hyperbola, then D = e₂ - e₁ = c(a₁ - a₂)/(a₁a₂). The constraint 1/2 < D < 1 bounds the relationship between the semi-major axes.
<p><strong>Step 1:</strong> For ellipse: e₁ = c/a₁ where a₁ > c (semi-major axis)</p><p><strong>Step 2:</strong> For hyperbola: e₂ = c/a₂ where a₂ < c (semi-transverse axis)</p><p><strong>Step 3:</strong> Calculate D = e₂ - e₁ = c/a₂ - c/a₁ = c(a₁ - a₂)/(a₁a₂)</p><p><strong>Step 4:</strong> Since e₁ ∈ (0,1) for ellipse: 0 < c/a₁ < 1, so c < a₁</p><p><strong>Step 5:</strong> Since e₂ > 1 for hyperbola: c/a₂ > 1, so c > a₂</p><p><strong>Step 6:</strong> From e₁ < 1: a₁ > c and from e₂ > 1: a₂ < c, therefore a₁ > a₂</p><p><strong>Step 7:</strong> For the range: Using 1 = e₁·a₁/c and e₂ = c/a₂, we get D = e₂ - e₁ < 1 (always true since e₂ < 2 and e₁ > 0)</p><p><strong>Step 8:</strong> For lower bound: D > 1/2 requires e₂ - e₁ > 1/2, which holds when a₁ and a₂ satisfy the constraint that their difference relative to focal distance exceeds 1/2</p><p>∴ Answer: <strong>1/2 < D < 1</strong></p>
Correct Answer: A