Ellipse
Locus Problems
Grade 11
Question:
<p>Let \(O(0, 0)\) and \(A(0, 1)\) be two fixed points. Then the locus of a point \(P\) such that the perimeter of \(\triangle AOP\) is 4, is:</p>
<p>\(8x^2 - 9y^2 + 9y = 18\)</p>
<p>\(9x^2 - 8y^2 + 8y = 16\)</p>
<p>\(9x^2 + 8y^2 - 8y = 16\)</p>
<p>\(8x^2 + 9y^2 - 9y = 18\)</p>
Step-by-Step Solution
Key Concept: The perimeter condition |OA| + |OP| + |PA| = 4 with fixed |OA| = 1 transforms to |OP| + |PA| = 3, which is the definition of an ellipse with foci at O and A where the sum of distances equals 3.
<p><strong>Step 1:</strong> Identify the fixed points: O(0, 0) and A(0, 1), so |OA| = 1.</p><p><strong>Step 2:</strong> The perimeter of △AOP is |OA| + |OP| + |PA| = 4.</p><p><strong>Step 3:</strong> Since |OA| = 1 is fixed, we have: |OP| + |PA| = 4 - 1 = 3.</p><p><strong>Step 4:</strong> By definition, the locus of points P such that the sum of distances to two fixed points (foci O and A) equals a constant is an <strong>ellipse</strong>.</p><p><strong>Step 5:</strong> Here, 2a = 3, so a = 3/2. The foci are at O(0, 0) and A(0, 1), so 2c = 1 and c = 1/2.</p><p><strong>Step 6:</strong> Using b² = a² - c² = (3/2)² - (1/2)² = 9/4 - 1/4 = 2.</p><p><strong>Step 7:</strong> The center of the ellipse is the midpoint of OA: (0, 1/2). The major axis is vertical.</p><p><strong>Step 8:</strong> The equation is: x² + (y - 1/2)²/(9/4) = 1, or equivalently 4x² + 16(y - 1/2)²/9 = 4.</p><p>∴ Answer: C</p>
Correct Answer: C