Ellipse
Definition and Eccentricity
Grade 11
Question:
<p>Find the eccentricity of an ellipse if the sum of distances from any point P on the ellipse to two foci F₁ at (-4, 4) and F₂ at (3, 3) equals \(7\sqrt{2}\), and the distance between foci is \(5\sqrt{2}\).</p>
Step-by-Step Solution
Key Concept: Use the definition of ellipse (sum of distances to foci = 2a) and the relationship between semi-major axis and focal distance to find eccentricity.
<p><strong>Given:</strong> P = (0, 0), F₁ = (-4, 4), F₂ = (3, 3)</p><p><strong>Step 1:</strong> Calculate PF₁ and PF₂:</p><p>\(PF_1 = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\)</p><p>\(PF_2 = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\)</p><p><strong>Step 2:</strong> Apply ellipse definition:</p><p>\(PF_1 + PF_2 = 2a\)</p><p>\(4\sqrt{2} + 3\sqrt{2} = 2a\)</p><p>\(7\sqrt{2} = 2a\)</p><p><strong>Step 3:</strong> Calculate distance between foci:</p><p>\(2ae = \sqrt{(3-(-4))^2 + (3-4)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}\)</p><p><strong>Step 4:</strong> Find eccentricity:</p><p>\(e = \frac{2ae}{2a} = \frac{5\sqrt{2}}{7\sqrt{2}} = \frac{5}{7}\)</p>
Correct Answer: c