Key Concepts and Formulas

• Equation of a Curve: A point $(x_0, y_0)$ lies on the curve $y = f(x)$ if and only if $y_0 = f(x_0)$.
• Slope of Tangent: The slope of the tangent to the curve $y = f(x)$ at the point $(x_0, y_0)$ is given by $m_T = \frac{dy}{dx}\Big|_{(x_0, y_0)}$.
• Slope of Normal: The normal to a curve at a point is perpendicular to the tangent at that point. Thus, $m_N = -\frac{1}{m_T}$, where $m_N$ is the slope of the normal.

Step-by-Step Solution

Step 1: Use the point (1, -3) to establish a relationship between a and b

Since the point (1, -3) lies on the curve $y = \frac{x - a}{(x + b)(x - 2)}$, we can substitute $x = 1$ and $y = -3$ into the equation: $-3 = \frac{1 - a}{(1 + b)(1 - 2)}$ $-3 = \frac{1 - a}{(1 + b)(-1)}$ Multiply both sides by $-(1 + b)$: $3(1 + b) = 1 - a$ $3 + 3b = 1 - a$ Rearrange the equation to isolate a: $a = -3b - 2 \quad \text{(Equation 1)}$ This equation provides the first relationship between the unknowns a and b.

Step 2: Find the slope of the normal and the tangent at (1, -3)

The equation of the normal is given as $x - 4y = 13$. We can rewrite this equation in slope-intercept form ($y = mx + c$) to find the slope of the normal: $4y = x - 13$ $y = \frac{1}{4}x - \frac{13}{4}$ Therefore, the slope of the normal is $m_N = \frac{1}{4}$.

Since the tangent is perpendicular to the normal, the slope of the tangent ($m_T$) is the negative reciprocal of the slope of the normal: $m_T = -\frac{1}{m_N} = -\frac{1}{\frac{1}{4}} = -4$ So, we know that $\frac{dy}{dx}\Big|_{(1, -3)} = -4$.

Step 3: Differentiate the curve equation to find $\frac{dy}{dx}$

The equation of the curve is $y = \frac{x - a}{(x + b)(x - 2)}$. We will use the quotient rule to differentiate this equation. Let $u = x - a$ and $v = (x + b)(x - 2)$. Then $u' = 1$. To find $v'$, we use the product rule: $v' = (1)(x - 2) + (x + b)(1) = x - 2 + x + b = 2x + b - 2$. Applying the quotient rule, $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$: $\frac{dy}{dx} = \frac{1 \cdot (x + b)(x - 2) - (x - a)(2x + b - 2)}{((x + b)(x - 2))^2}$

Step 4: Evaluate the derivative at x = 1

Substitute $x = 1$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx}\Big|_{x=1} = \frac{(1 + b)(1 - 2) - (1 - a)(2(1) + b - 2)}{((1 + b)(1 - 2))^2}$ $\frac{dy}{dx}\Big|_{x=1} = \frac{(1 + b)(-1) - (1 - a)(b)}{((1 + b)(-1))^2}$ $\frac{dy}{dx}\Big|_{x=1} = \frac{-(1 + b) - b(1 - a)}{(1 + b)^2}$

Step 5: Equate the slope of the tangent and solve for b

We know that $\frac{dy}{dx}\Big|_{x=1} = -4$. Therefore: $\frac{-(1 + b) - b(1 - a)}{(1 + b)^2} = -4$ Multiply both sides by $(1 + b)^2$: $-(1 + b) - b(1 - a) = -4(1 + b)^2$ $-1 - b - b + ab = -4(1 + 2b + b^2)$ $-1 - 2b + ab = -4 - 8b - 4b^2$

Now, substitute $a = -3b - 2$ from Equation 1 into this equation: $-1 - 2b + (-3b - 2)b = -4 - 8b - 4b^2$ $-1 - 2b - 3b^2 - 2b = -4 - 8b - 4b^2$ $-3b^2 - 4b - 1 = -4b^2 - 8b - 4$ Move all terms to one side to form a quadratic equation in b: $b^2 + 4b + 3 = 0$ Factor the quadratic equation: $(b + 1)(b + 3) = 0$ This gives two possible values for b: $b = -1 \quad \text{or} \quad b = -3$

However, if $b = -1$, the original equation becomes $y = \frac{x - a}{(x - 1)(x - 2)}$, and the derivative at $x=1$ would be undefined. Therefore, $b$ cannot be $-1$. Thus, $b = -3$.

Step 6: Solve for a

Substitute $b = -3$ into Equation 1: $a = -3(-3) - 2$ $a = 9 - 2$ $a = 7$

Step 7: Calculate a + b

$a + b = 7 + (-3) = 4$

Step 8: Check for mistakes

We need to re-evaluate the derivative with the found values of a and b, to see if we made a mistake.

$y = \frac{x-7}{(x-3)(x-2)} = \frac{x-7}{x^2 - 5x + 6}$

$\frac{dy}{dx} = \frac{1*(x^2 - 5x + 6) - (x-7)(2x-5)}{(x^2 - 5x + 6)^2} = \frac{x^2 - 5x + 6 - (2x^2 - 5x - 14x + 35)}{(x^2 - 5x + 6)^2} = \frac{-x^2 + 14x - 29}{(x^2 - 5x + 6)^2}$

At $x=1$, $\frac{dy}{dx} = \frac{-1 + 14 - 29}{(1 - 5 + 6)^2} = \frac{-16}{4} = -4$.

This is correct.

Common Mistakes & Tips

• Quotient Rule: Be extra careful when applying the quotient rule to avoid sign errors.
• Checking for Extraneous Solutions: Always check if the solutions obtained are valid by substituting them back into the original equation. In this case, we had to discard $b = -1$ because it made the derivative undefined at $x=1$.
• Simplifying Expressions: Simplifying expressions after each step can help reduce the chance of making errors.

Summary

We started by using the point (1, -3) lying on the curve to establish a relationship between a and b. Then, we found the slope of the tangent using the equation of the normal. We differentiated the curve equation and equated it to the slope of the tangent. Finally, we solved for a and b and calculated $a + b$.

The final answer is \boxed{4}.