Ellipse
Relation between Ellipse and Hyperbola
Grade 11
Question:
<p>An ellipse passes through the foci of the hyperbola, \(9x^2 - 4y^2 = 36\) and its major and minor axes lie along the transverse and conjugate axes of the hyperbola, respectively. If the product of eccentricities of the two conics is \(1/2\), then which of the following points does not lie on the ellipse?</p>
<p>\((\sqrt{13}, 0)\)</p>
<p>\(\left(\dfrac{\sqrt{39}}{2}, \sqrt{3}\right)\)</p>
<p>\(\left(\dfrac{1}{2}\sqrt{13}, \dfrac{\sqrt{3}}{2}\right)\)</p>
<p>\(\left(\dfrac{\sqrt{13}}{2}, \sqrt{6}\right)\)</p>
Step-by-Step Solution
Key Concept: First find hyperbola's foci using its eccentricity, then use the condition that the ellipse passes through these foci and the eccentricity product constraint to determine the ellipse equation.
<p><strong>Step 1: Standardize and find hyperbola parameters</strong></p><p>Hyperbola: 9x² - 4y² = 36 → x²/4 - y²/9 = 1</p><p>Here a₁² = 4, b₁² = 9, so a₁ = 2, b₁ = 3</p><p>For hyperbola: c₁² = a₁² + b₁² = 4 + 9 = 13, so c₁ = √13</p><p>Eccentricity: e₁ = c₁/a₁ = √13/2</p><p><strong>Step 2: Identify hyperbola's foci</strong></p><p>Foci of hyperbola: (±√13, 0) [on transverse axis]</p><p><strong>Step 3: Set up ellipse equation</strong></p><p>Ellipse has major axis along x-axis, minor axis along y-axis: x²/A² + y²/B² = 1 where A > B</p><p>Ellipse passes through (√13, 0): 13/A² = 1, so A² = 13</p><p><strong>Step 4: Use eccentricity product condition</strong></p><p>For ellipse: e₂ = √(1 - B²/A²) = √(1 - B²/13)</p><p>Given: e₁ · e₂ = 1/2</p><p>(√13/2) · √(1 - B²/13) = 1/2</p><p>√(1 - B²/13) = 1/√13</p><p>1 - B²/13 = 1/13</p><p>B²/13 = 12/13, so B² = 12</p><p><strong>Step 5: Ellipse equation</strong></p><p>x²/13 + y²/12 = 1</p><p>Test given options by substituting into this equation. The point that does NOT satisfy the equation is the answer.</p><p>∴ Answer: C</p>
Correct Answer: C