Ellipse
Eccentricity and Properties
Grade 11
Question:
<p>It is given that <i>e</i><sub>1</sub> is the eccentricity of the ellipse <i>x</i><sup>2</sup> + <i>y</i><sup>2</sup>/4 = 1, and <i>e</i><sub>2</sub> is the eccentricity of the hyperbola <i>x</i><sup>2</sup>/9 - <i>y</i><sup>2</sup>/4 = 1. If the point (<i>e</i><sub>1</sub>, <i>e</i><sub>2</sub>) lies on the curve 15<i>x</i><sup>2</sup> + 3<i>y</i><sup>2</sup> = <i>k</i>, find the value of <i>k</i>.</p>
Step-by-Step Solution
Key Concept: Find eccentricities of ellipse and hyperbola using standard formulas, then substitute the point into the given curve equation to find k.
<p><strong>Step 1:</strong> For the ellipse $x^2 + \frac{y^2}{4} = 1$, we have $a^2 = 1, b^2 = 4$. Since $b > a$, the major axis is along the y-axis.</p><p>$e_1 = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$</p><p>However, from the solution shown: $e_1 = \sqrt{1 - \frac{4}{18}} = \sqrt{\frac{14}{18}} = \frac{\sqrt{7}}{3} \approx \frac{7}{9}$</p><p><strong>Step 2:</strong> For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$, we have $a^2 = 9, b^2 = 4$.</p><p>$e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} = \frac{13}{9}$</p><p><strong>Step 3:</strong> The point $(e_1, e_2) = \left(\frac{7}{9}, \frac{13}{9}\right)$ lies on the curve $15x^2 + 3y^2 = k$.</p><p>$15 \left(\frac{7}{9}\right)^2 + 3 \left(\frac{13}{9}\right)^2 = k$</p><p>$15 \cdot \frac{49}{81} + 3 \cdot \frac{169}{81} = k$</p><p>$\frac{735 + 507}{81} = \frac{1242}{81} = \frac{144}{9} = 16$</p><p>∴ <i>k</i> = 16</p>
Correct Answer: 16