Key Concepts and Formulas

• Equation of a Tangent Line: The equation of the tangent to a curve $y=f(x)$ at a point $(x_0, y_0)$ is given by $y - y_0 = f'(x_0)(x - x_0)$.
• Derivative of a Quotient: The derivative of a function $\frac{u(x)}{v(x)}$ is given by $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$. Recognizing the form $\frac{x f'(x) - f(x)}{x^2}$ as the derivative of $\frac{f(x)}{x}$ is crucial.
• Distance Formula: The square of the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.

Step-by-Step Solution

Step 1: Formulate the Equation of the Tangent at Point R We are given the curve $y=f(x)$. The point $R$ is $(b, f(b))$. Our goal is to find the equation of the tangent line at $R$. The slope of the tangent at $R$ is $f'(b)$. Using the point-slope form, the equation of the tangent line at $R(b, f(b))$ is: $y - f(b) = f'(b)(x - b)$

Step 2: Use the y-intercept Condition The tangent line cuts the y-axis at point $S(0, c)$. This means that when $x=0$, $y=c$. We substitute $x=0$ and $y=c$ into the tangent equation to relate $c$, $f(b)$, and $f'(b)$. $c - f(b) = f'(b)(0 - b)$ $c - f(b) = -b f'(b)$ Rearranging this equation, we get: $b f'(b) - f(b) = -c$

Step 3: Incorporate the Given Condition $bc=3$ We are given the condition $bc=3$. This allows us to express $c$ in terms of $b$: $c = \frac{3}{b}$. We substitute this expression for $c$ into the equation from Step 2. $b f'(b) - f(b) = -\frac{3}{b}$

Step 4: Transform into a Recognizable Differential Equation The left side of the equation, $b f'(b) - f(b)$, resembles the numerator of the quotient rule derivative formula $\frac{d}{dx}\left(\frac{f(x)}{x}\right) = \frac{x f'(x) - f(x)}{x^2}$. To match this form, we divide both sides of our equation by $b^2$. $\frac{b f'(b) - f(b)}{b^2} = \frac{-\frac{3}{b}}{b^2}$ $\frac{b f'(b) - f(b)}{b^2} = -\frac{3}{b^3}$ Now, the left side is exactly the derivative of $\frac{f(b)}{b}$ with respect to $b$: $\frac{d}{db}\left(\frac{f(b)}{b}\right) = -\frac{3}{b^3}$

Step 5: Integrate to Find the General Form of $f(x)$ We integrate both sides of the differential equation with respect to $b$ to find the general form of $f(x)$. $\int \frac{d}{db}\left(\frac{f(b)}{b}\right) db = \int -\frac{3}{b^3} db$ $\frac{f(b)}{b} = -3 \int b^{-3} db$ $\frac{f(b)}{b} = -3 \left( \frac{b^{-2}}{-2} \right) + C_1$ $\frac{f(b)}{b} = \frac{3}{2b^2} + C_1$ Multiplying by $b$ to solve for $f(b)$: $f(b) = \frac{3}{2b} + C_1 b$ Since the curve is $y=f(x)$, we can replace $b$ with $x$: $f(x) = \frac{3}{2x} + C_1 x$

Step 6: Use Point P to Find the Constant of Integration ($C_1$) The curve passes through the point $P\left(1, \frac{3}{2}\right)$. We substitute $x=1$ and $f(x)=\frac{3}{2}$ into our equation for $f(x)$ to find $C_1$. $\frac{3}{2} = \frac{3}{2(1)} + C_1 (1)$ $\frac{3}{2} = \frac{3}{2} + C_1$ $C_1 = 0$

Step 7: Determine the Specific Function $f(x)$ We substitute $C_1=0$ back into the expression for $f(x)$ to find the specific function. $f(x) = \frac{3}{2x} + (0)x$ $f(x) = \frac{3}{2x}$

Step 8: Use Point Q to Find the Value of 'a' The curve also passes through the point $Q\left(a, \frac{1}{2}\right)$. We substitute $x=a$ and $f(x)=\frac{1}{2}$ into the specific function $f(x)$ to find $a$. $\frac{1}{2} = \frac{3}{2a}$ To solve for $a$, cross-multiply: $2a = 6$ $a = 3$

Step 9: Calculate $(PQ)^2$ Now we have the coordinates of both points: $P = \left(1, \frac{3}{2}\right)$ $Q = \left(3, \frac{1}{2}\right)$ Using the distance formula, $(PQ)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$: $(PQ)^2 = (3 - 1)^2 + \left(\frac{1}{2} - \frac{3}{2}\right)^2$ $(PQ)^2 = (2)^2 + \left(-\frac{2}{2}\right)^2$ $(PQ)^2 = 4 + (-1)^2$ $(PQ)^2 = 4 + 1$ $(PQ)^2 = 5$

Common Mistakes & Tips

• Recognizing the Differential Form: The most critical step is recognizing that $\frac{b f'(b) - f(b)}{b^2}$ is the derivative of $\frac{f(b)}{b}$.
• Constant of Integration: Don't forget the constant of integration ($C_1$) after integrating.
• Careful with signs: A simple mistake with signs will lead to a wrong answer.

Summary

This problem integrates concepts from differential calculus, integral calculus, and coordinate geometry. The key is to translate the given geometric condition into a differential equation, recognize a standard differential form, solve the differential equation, and then use the given points to find the specific function and the value of $a$. Finally, use the distance formula to calculate $(PQ)^2$.

The final answer is $\boxed{5}$.