A circle $S$ of radius unity touches a line $L$ at $P$. A point $A$ lies on $S$ and $N$ is the foot of the perpendicular from $A$ to $L$. The area of $\triangle PAN$ as $A$ varies cannot be equal to:
Step-by-Step Solution
Key Concept: The maximum area is found by expressing it as a function of the parametric angle and using calculus to locate critical points.
Given circle $S: x^2 + (y-1)^2 = 1$ touching line $L$ (x-axis) at $P(0,0)$, and point $A(\cos\theta, 1+\sin\theta)$ on the circle. Area of $\triangle PAN$ is $\frac{1}{2}|\cos\theta(1+\sin\theta)| = \frac{1}{2}|f(\theta)|$ where $f(\theta) = \cos\theta(1+\sin\theta)$. Taking derivative: $f'(\theta) = \cos 2\theta - \sin\theta = 0$ yields $\theta = \pi/6, 5\pi/6, 3\pi/2$. Checking second derivative shows maxima at $\theta = \pi/6$ and $5\pi/6$ with $f(\pi/6) = f(5\pi/6) = 3\sqrt{3}/4$. Maximum area of $\triangle PAN$ is $\boxed{3\sqrt{3}/8}$.
Correct Answer: 1,2,4