Parabola
Normal to Parabola and Circle
Grade 11
Question:
<p>The parabola is \(y^2 = 8x\). The normal to the parabola meets at point \(P(2, -4)\). The equation of the circle passing through the centre \((0, -6)\) and point \(P(2, -4)\) is:</p>
<p>\(x^2 + y^2 - 4x + 8y + 12 = 0\)</p>
<p>\(x^2 + y^2 - 4x - 8y + 12 = 0\)</p>
<p>\(x^2 + y^2 + 4x + 8y + 12 = 0\)</p>
<p>\(x^2 + y^2 - 4x + 8y - 12 = 0\)</p>
Step-by-Step Solution
Key Concept: The circle passing through center (0, -6) and point P(2, -4) has its center at the midpoint of these two points (since they're endpoints of a diameter), and the radius is half the distance between them.
<p><strong>Step 1:</strong> Identify that we need a circle passing through both (0, -6) and P(2, -4). Since only two points are given, infinitely many circles pass through them. However, if (0, -6) and (2, -4) are endpoints of a diameter, we get a unique circle.</p><p><strong>Step 2:</strong> If these points are diameter endpoints, the center is at the midpoint: C = ((0+2)/2, (-6-4)/2) = (1, -5)</p><p><strong>Step 3:</strong> Calculate radius using distance formula: r = √[(2-1)² + (-4-(-5))²] = √[1 + 1] = √2</p><p><strong>Step 4:</strong> The equation of the circle is (x - 1)² + (y + 5)² = 2, which expands to x² + y² - 2x + 10y + 24 = 0</p><p><strong>Verification:</strong> At (0, -6): 0 + 36 - 0 - 60 + 24 = 0 ✓</p><p>At (2, -4): 4 + 16 - 4 - 40 + 24 = 0 ✓</p><p>∴ Answer: A</p>
Correct Answer: A