Ellipse
Equation of Ellipse from foci and eccentricity
Grade 11
Question:
<p>Equation of foci of an ellipse are \((\pm 2, 0)\) and \(e = \frac{1}{2}\). Then the equation of the ellipse is:</p>
<p>\(\frac{x^2}{16} + \frac{y^2}{12} = 1\)</p>
<p>\(\frac{x^2}{12} + \frac{y^2}{16} = 1\)</p>
<p>\(\frac{x^2}{16} + \frac{y^2}{4} = 1\)</p>
<p>\(\frac{x^2}{4} + \frac{y^2}{16} = 1\)</p>
Step-by-Step Solution
Key Concept: Use the relationship c = ae where c is the distance from center to focus, a is semi-major axis, and e is eccentricity. From foci at (±2, 0), we get c = 2, so 2 = a(1/2) gives a = 4. Then b² = a² - c² = 16 - 4 = 12.
<p><strong>Step 1:</strong> Identify given information. Foci are at (±2, 0), so foci lie on x-axis, meaning major axis is along x-axis. The distance from center to each focus is c = 2.</p><p><strong>Step 2:</strong> Use eccentricity relation e = c/a. Given e = 1/2 and c = 2:<br/>1/2 = 2/a<br/>∴ a = 4</p><p><strong>Step 3:</strong> Find b using b² = a² - c²:<br/>b² = 16 - 4 = 12<br/>∴ b = 2√3</p><p><strong>Step 4:</strong> Write standard form with major axis along x-axis:<br/>x²/a² + y²/b² = 1<br/>x²/16 + y²/12 = 1</p><p>∴ Answer: A</p>
Correct Answer: A