<p>If \(y = mx + c\) is the normal at a point on the parabola \(y^2 = 8x\) whose focal distance is 8 units then \(|c|\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the focal distance property of parabola (distance from point to focus equals distance to directrix) to find the point, then apply the normal equation condition to determine the intercept.
<p><strong>Step 1:</strong> For parabola y² = 8x, we have 4a = 8, so a = 2. Focus is at (2, 0).</p><p><strong>Step 2:</strong> For a point P(t², 2t) on the parabola, focal distance = t² + a = t² + 2 = 8. Thus t² = 6, giving t = ±√6.</p><p><strong>Step 3:</strong> The normal at point (t², 2t) on y² = 4ax is: y = -tx + 2t + t³.</p><p><strong>Step 4:</strong> Substituting t² = 6: Normal equation is y = -tx + 2t + t·t² = -tx + 2t + 6t.</p><p>This gives y = -tx + 8t, where t = ±√6.</p><p><strong>Step 5:</strong> Comparing with y = mx + c: c = 8t = ±8√6.</p><p><strong>Step 6:</strong> Therefore |c| = 8√6.</p><p>∴ Answer: D</p>
Correct Answer: D