Ellipse
Definition and Standard Form
Grade 11
Question:
<p>Show that the equation of the locus of a point which moves so that the sum of its distances from two given points \((ae, 0)\) and \((-ae, 0)\) is equal to \(2a\), is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(b^2 = a^2(1 - e^2)\).</p>
Step-by-Step Solution
Key Concept: Use the definition of an ellipse: the locus of points where the sum of distances to two fixed foci equals a constant 2a. Algebraically eliminate the square roots by isolating and squaring twice, then simplify using the relationship b² = a²(1-e²).
<p><strong>Step 1:</strong> Let P(x, y) be a point on the locus. The two fixed points (foci) are F₁(ae, 0) and F₂(-ae, 0).</p><p>Given condition: PF₁ + PF₂ = 2a</p><p>∴ √[(x-ae)² + y²] + √[(x+ae)² + y²] = 2a</p><p><strong>Step 2:</strong> Let √[(x-ae)² + y²] = r₁. Then √[(x+ae)² + y²] = 2a - r₁</p><p>Squaring: (x+ae)² + y² = 4a² - 4ar₁ + r₁²</p><p>(x+ae)² + y² = 4a² - 4a√[(x-ae)² + y²] + (x-ae)² + y²</p><p><strong>Step 3:</strong> Simplify: (x+ae)² - (x-ae)² = 4a² - 4a√[(x-ae)² + y²]</p><p>4aex = 4a² - 4a√[(x-ae)² + y²]</p><p>∴ √[(x-ae)² + y²] = a - ex</p><p><strong>Step 4:</strong> Square again: (x-ae)² + y² = a² - 2aex + e²x²</p><p>x² - 2aex + a²e² + y² = a² - 2aex + e²x²</p><p>x² + y² + a²e² = a² + e²x²</p><p>x²(1-e²) + y² = a²(1-e²)</p><p><strong>Step 5:</strong> Divide by a²(1-e²). Since b² = a²(1-e²):</p><p>x²/a² + y²/b² = 1</p><p>∴ <strong>Answer: The locus is the ellipse x²/a² + y²/b² = 1 where b² = a²(1-e²)</strong></p>
Correct Answer: Locus: x²/a² + y²/b² = 1 where b² = a²(1-e²)