Ellipse
Equation of Ellipse
Grade 11
Question:
<p>An ellipse has eccentricity \(e = \dfrac{2}{5}\) (i.e., \(1 - e^2 = \dfrac{3}{5}\)) and passes through the point \\(\left(\dfrac{9}{a^2} + \dfrac{5}{3a^2} = 1\right)\). Find the equation of the ellipse.</p>
<p>\(3x^2 + 5y^2 - 32 = 0\)</p>
<p>\(5x^2 + 3y^2 - 32 = 0\)</p>
<p>\(3x^2 + 5y^2 + 32 = 0\)</p>
<p>\(5x^2 + 3y^2 + 32 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the eccentricity relation b²/a² = 1 - e² to express b in terms of a, then substitute the given point into the ellipse equation x²/a² + y²/b² = 1 to find the specific value of a.
<p><strong>Step 1:</strong> Use the eccentricity relation. Given e = 2/5, we have e² = 4/25, so 1 - e² = 21/25. But we're told 1 - e² = 3/5 = 15/25, which gives b²/a² = 3/5, or <strong>b² = (3/5)a²</strong>.</p><p><strong>Step 2:</strong> The ellipse equation is x²/a² + y²/b² = 1. The point lies on the ellipse. Interpreting the condition given as the point satisfying the ellipse equation: substitute b² = (3/5)a² to get x²/a² + 5y²/(3a²) = 1.</p><p><strong>Step 3:</strong> From the problem structure (9/a² + 5/(3a²) = 1), this indicates x² = 9 and y² = 1 (or similar values). Testing: 9/a² + 5·1/(3a²) = 9/a² + 5/(3a²) = (27 + 5)/(3a²) = 32/(3a²) = 1, giving <strong>a² = 32/3</strong>.</p><p><strong>Step 4:</strong> Then b² = (3/5)·(32/3) = 32/5. The ellipse equation is <strong>3x²/32 + 5y²/32 = 1</strong> or equivalently <strong>3x² + 5y² = 32</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A