Problem Analysis #0

To solve the problem, we analyze the conditions given and derive the constraints on the parabola $C: y = ax^2 + bx + c$.

Step 1: Tangent Condition at $P$ and $Q$
The tangent to the unit circle at $P(\cos\theta, \sin\theta)$ is $x \cos\theta + y \sin\theta = 1$. For $C$ to share this tangent at $P$, the slopes must match.

• The derivative of $C$ is $y' = 2ax + b$.
• At $P$, the slope of the tangent to $C$ is $2a \cos\theta + b$.
• The slope of the tangent to the circle at $P$ is $-\cot\theta$.
  Equating these gives:

$$2a \cos\theta + b = -\cot\theta \quad \Rightarrow \quad b = -2a \cos\theta - \cot\theta.$$

Step 2: Tangent Condition at $Q$
Similarly, for $Q(-\cos\theta, \sin\theta)$, the tangent is $x \cos\theta - y \sin\theta = 1$.

• The slope of the tangent to $C$ at $Q$ is $2a(-\cos\theta) + b = -2a \cos\theta + b$.
• The slope of the tangent to the circle at $Q$ is $\cot\theta$.
  Equating these gives:

$$-2a \cos\theta + b = \cot\theta \quad \Rightarrow \quad b = 2a \cos\theta + \cot\theta.$$

Step 3: Solving for $a$ and $b$
Equating the two expressions for $b$:

$$-2a \cos\theta - \cot\theta = 2a \cos\theta + \cot\theta \quad \Rightarrow \quad -4a \cos\theta = 2 \cot\theta \quad \Rightarrow \quad a = -\frac{\cot\theta}{2\cos\theta} = -\frac{1}{2\sin\theta}.$$

Substituting $a$ back into $b = -2a \cos\theta - \cot\theta$:

$$b = -2\left(-\frac{1}{2\sin\theta}\right)\cos\theta - \cot\theta = \frac{\cos\theta}{\sin\theta} - \cot\theta = 0.$$

Thus, $b = 0$.

Step 4: Finding $c$
Since $C$ passes through $P(\cos\theta, \sin\theta)$:

$$\sin\theta = a \cos^2\theta + b \cos\theta + c \quad \Rightarrow \quad c = \sin\theta - a \cos^2\theta.$$

Substituting $a = -\frac{1}{2\sin\theta}$:

$$c = \sin\theta - \left(-\frac{1}{2\sin\theta}\right)\cos^2\theta = \sin\theta + \frac{\cos^2\theta}{2\sin\theta}.$$

Step 5: Area Calculation
The area $A$ under the parabola from $-1$ to $1$ is:

$$A = \int_{-1}^{1} \left(-\frac{1}{2\sin\theta}x^2 + \frac{\cos^2\theta}{2\sin\theta}\right) dx.$$

Evaluating the integral:

$$A = \left[-\frac{1}{6\sin\theta}x^3 + \frac{\cos^2\theta}{2\sin\theta}x\right]_{-1}^{1} = \frac{2}{3\sin\theta} + \frac{\cos^2\theta}{\sin\theta}.$$

Simplifying:

$$A = \frac{2 + 3\cos^2\theta}{3\sin\theta}.$$

Thus, $A^2 = \left(\frac{2 + 3\cos^2\theta}{3\sin\theta}\right)^2$.

Step 6: Minimizing $A^2$
To minimize $A^2$, we need to minimize $\frac{2 + 3\cos^2\theta}{3\sin\theta}$.
Using trigonometric identities and calculus, we find that the minimum occurs when $\theta = 45^\circ$, yielding:

$$A^2 = \left(\frac{2 + 3 \cdot \frac{1}{2}}{3 \cdot \frac{\sqrt{2}}{2}}\right)^2 = \left(\frac{5/2}{3\sqrt{2}/2}\right)^2 = \left(\frac{5}{3\sqrt{2}}\right)^2 = \frac{25}{18}.$$

Final Answer

$$\boxed{\dfrac{25}{18}}$$