Ellipse
Eccentricity from Foci
Grade 11
Question:
<p>An ellipse having foci at (3, 3) and (-4, 4) and passing through the origin has eccentricity equal to:</p>
<p>(a) \(\frac{3}{7}\)</p>
<p>(b) \(\frac{2}{7}\)</p>
<p>(c) \(\frac{5}{7}\)</p>
<p>(d) \(\frac{3}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the focal definition of an ellipse: sum of distances from any point to the two foci equals 2a. Calculate distances from origin to both foci.
<p>The foci are \(F_1(3, 3)\) and \(F_2(-4, 4)\). Distance between foci: \(2c = \sqrt{(3-(-4))^2 + (3-4)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}\), so \(c = \frac{5\sqrt{2}}{2}\). Since the ellipse passes through origin (0,0), use the focal property: \(|F_1O| + |F_2O| = 2a\). \(|F_1O| = \sqrt{9+9} = 3\sqrt{2}\); \(|F_2O| = \sqrt{16+16} = 4\sqrt{2}\). Thus \(2a = 7\sqrt{2}\), so \(a = \frac{7\sqrt{2}}{2}\). Eccentricity: \(e = \frac{c}{a} = \frac{5\sqrt{2}/2}{7\sqrt{2}/2} = \frac{5}{7}\).</p>
Correct Answer: C