<p>P is a variable point on the curve <i>y</i> = <i>f</i>(<i>x</i>) and A is a fixed point in the plane not lying on the curve. If PA is minimum, then the angle between PA and the tangent at P is</p>
Step-by-Step Solution
Key Concept: When PA (distance from point P on curve to fixed point A) is minimized, the line PA must be perpendicular to the tangent at P. This is because at the minimum distance, any infinitesimal movement along the curve should not decrease the distance further, which occurs only when PA is normal to the curve.
<p><strong>Step 1:</strong> Let P(x, f(x)) be a variable point on the curve y = f(x), and A(a, b) be the fixed point.</p><p><strong>Step 2:</strong> The distance PA is given by: D² = (x - a)² + (f(x) - b)²</p><p><strong>Step 3:</strong> For PA to be minimum, we take the derivative with respect to x and set it equal to zero:</p><p>d(D²)/dx = 2(x - a) + 2(f(x) - b)·f'(x) = 0</p><p><strong>Step 4:</strong> This gives us: (x - a) + (f(x) - b)·f'(x) = 0</p><p><strong>Step 5:</strong> Rearranging: (x - a) = -(f(x) - b)·f'(x)</p><p><strong>Step 6:</strong> The slope of PA is: m₁ = (f(x) - b)/(x - a)</p><p><strong>Step 7:</strong> The slope of the tangent at P is: m₂ = f'(x)</p><p><strong>Step 8:</strong> From Step 4, we have: (x - a) = -(f(x) - b)·f'(x), which means:</p><p>m₁ × m₂ = [(f(x) - b)/(x - a)] × f'(x) = -1</p><p><strong>Step 9:</strong> This condition m₁ × m₂ = -1 proves that PA and the tangent at P are perpendicular, i.e., the angle between them is π/2.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C