Ellipse
Eccentricity
Grade 11
Question:
<p>If <i>e</i><sub>1</sub> and <i>e</i><sub>2</sub> are the eccentricities of the ellipse, \(\frac{x^2}{18} + \frac{y^2}{4} = 1\) and the hyperbola, \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) respectively and (<i>e</i><sub>1</sub>, <i>e</i><sub>2</sub>) is a point on the ellipse, \(15x^2 + 3y^2 = k\), then <i>k</i> is equal to</p>
<p>(a) 14</p>
<p>(b) 15</p>
<p>(c) 17</p>
<p>(d) 16</p>
Step-by-Step Solution
Key Concept: Calculate eccentricities of ellipse and hyperbola using their standard formulas, then substitute the point into the given equation.
<p><strong>Step 1:</strong> Find eccentricity <i>e</i><sub>1</sub> of the ellipse \(\frac{x^2}{18} + \frac{y^2}{4} = 1\).</p><p>Here, \(a^2 = 18\), \(b^2 = 4\), so \(e_1^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{4}{18} = \frac{14}{18} = \frac{7}{9}\)</p><p>Therefore, \(e_1 = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}\)</p><p><strong>Step 2:</strong> Find eccentricity <i>e</i><sub>2</sub> of the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\).</p><p>Here, \(a^2 = 9\), \(b^2 = 4\), so \(e_2^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{4}{9} = \frac{13}{9}\)</p><p>Therefore, \(e_2 = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}\)</p><p><strong>Step 3:</strong> Point (<i>e</i><sub>1</sub>, <i>e</i><sub>2</sub>) lies on \(15x^2 + 3y^2 = k\).</p><p>Substituting: \(k = 15e_1^2 + 3e_2^2 = 15 \cdot \frac{7}{9} + 3 \cdot \frac{13}{9} = \frac{105}{9} + \frac{39}{9} = \frac{144}{9} = 16\)</p><p>∴ Answer is <strong>(d) 16</strong>.</p>
Correct Answer: D