Problem Analysis #0

To find the minimum possible value of $A^2$, we first determine the equation of the parabola $C: y = ax^2 + bx + c$.

Since the parabola passes through the symmetric points $P(\cos\theta, \sin\theta)$ and $Q(-\cos\theta, \sin\theta)$, it must be symmetric with respect to the $y$-axis. Thus, $b = 0$, and the equation simplifies to:

$$y = ax^2 + c$$

Since $P(\cos\theta, \sin\theta)$ lies on the parabola, we have:

$$\sin\theta = a\cos^2\theta + c \tag{1}$$

The unit circle is given by $x^2 + y^2 = 1$. The slope of the tangent to the circle at $P(\cos\theta, \sin\theta)$ is:

$$m = -\frac{\cos\theta}{\sin\theta} = -\cot\theta$$

The derivative of the parabola is $y' = 2ax$. At $P$, the slope of the tangent to the parabola is 2a\cos$$\theta$$$. Since the parabola and the circle share a common tangent at P$, their slopes must be equal:

$$2a\cos\theta = -\frac{\cos\theta}{\sin\theta}$$

Since 0^\circ < $$\theta$$ < 90^\circ, we have $\cos\theta \neq 0$. Dividing both sides by $\cos\theta$ gives:

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

Substituting $a$ back into equation (1) to find $c$:

$$\sin\theta = -\frac{\cos^2\theta}{2\sin\theta} + c \implies c = \sin\theta + \frac{\cos^2\theta}{2\sin\theta} = \frac{\sin^2\theta + 1}{2\sin\theta}$$

Let $t = \sin\theta$. Since 0^\circ < $$\theta$$ < 90^\circ, we have $t \in (0, 1)$. The equation of the parabola is:

$$y = -\frac{1}{2t}x^2 + \frac{t^2+1}{2t}$$

The $x$-intercepts of the parabola are found by setting $y = 0$:

$$x^2 = t^2 + 1 \implies x = \pm\sqrt{t^2+1}$$

The area $A$ of the region enclosed by the parabola and the $x$-axis is given by:

$$A = \frac{2}{3} \cdot \text{width} \cdot \text{height} = \frac{2}{3} \left(2\sqrt{t^2+1}\right) \left(\frac{t^2+1}{2t}\right) = \frac{2}{3t}(t^2+1)^{3/2}$$

We want to minimize $A^2$:

$$A^2 = \frac{4}{9t^2}(t^2+1)^3$$

Let $u = t^2$. Since $t \in (0, 1)$, we have $u \in (0, 1)$. We rewrite $A^2$ as:

$$A^2 = \frac{4}{9} \frac{(u+1)^3}{u}$$

To find the minimum of $g(u) = \frac{(u+1)^3}{u}$ for $u \in (0, 1)$, we take the derivative with respect to $u$:

$$g'(u) = \frac{3(u+1)^2 u - (u+1)^3}{u^2} = \frac{(u+1)^2(2u-1)}{u^2}$$

Setting $g'(u) = 0$ gives $u = \frac{1}{2}$. Since $g'(u) < 0$ for $u < \frac{1}{2}$ and $g'(u) > 0$ for $u > \frac{1}{2}$, the minimum occurs at $u = \frac{1}{2}$.

Evaluating $g\left(\frac{1}{2}\right)$:

$$g\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}+1\right)^3}{\frac{1}{2}} = 2 \left(\frac{3}{2}\right)^3 = \frac{27}{4}$$

Thus, the minimum value of $A^2$ is:

$$A^2 = \frac{4}{9} \cdot \frac{27}{4} = 3$$

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