Question:

Step 1: Use the given point on the curve to form an equation. The point $(1,-3)$ lies on the curve $y=\dfrac{x-a}{(x+b)(x-2)}$. Substitute $x=1$ and $y=-3$ into the curve's equation. $$ -3 = \dfrac{1-a}{(1+b)(1-2)} $$ $$ -3 = \dfrac{1-a}{(1+b)(-1)} $$ $$ 3(1+b) = 1-a $$ $$ 3+3b = 1-a $$ $$ a+3b = -2 \quad \text{(Equation 1)} $$ Step 2: Determine the slope of the normal from its equation. The equation of the normal to the curve at $(1,-3)$ is given as $x-4y=13$. We can rewrite this in the slope-intercept form $y=mx+c$. $$ 4y = x-13 $$ $$ y = \dfrac{1}{4}x - \dfrac{13}{4} $$ The slope of the normal, $m_N$, is $m_N = \dfrac{1}{4}$. Step 3: Determine the slope of the tangent at the given point. The slope of the tangent, $m_T$, at the point of contact is the negative reciprocal of the slope of the normal. $$ m_T = -\dfrac{1}{m_N} = -\dfrac{1}{1/4} = -4 $$ Thus, $\left(\dfrac{dy}{dx}\right)_{(1,-3)} = -4$. Step 4: Calculate the derivative of the curve and set it equal to the tangent's slope. The equation of the curve is $y=\dfrac{x-a}{(x+b)(x-2)} = \dfrac{x-a}{x^2+(b-2)x-2b}$. Using the quotient rule $\dfrac{dy}{dx} = \dfrac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, where $u=x-a$ and $v=x^2+(b-2)x-2b$: $$ \dfrac{dy}{dx} = \dfrac{(x^2+(b-2)x-2b)(1) - (x-a)(2x+b-2)}{(x^2+(b-2)x-2b)^2} $$ Now, substitute $x=1$ into the derivative expression. The value of $y=-3$ is implicitly used by the point being on the curve, which led to Equation 1. The denominator term at $x=1$ is $(1^2+(b-2)(1)-2b)^2 = (1+b-2-2b)^2 = (-1-b)^2 = (1+b)^2$. The numerator term at $x=1$ is $(1+(b-2)-2b) - (1-a)(2(1)+b-2)$ $$ = (1+b-2-2b) - (1-a)(2+b-2) $$ $$ = (-1-b) - (1-a)(b) $$ $$ = -1-b - b + ab $$ $$ = -1-2b+ab $$ So, $\left(\dfrac{dy}{dx}\right)_{(1,-3)} = \dfrac{-1-2b+ab}{(1+b)^2}$. Equating this to the slope of the tangent found in Step 3: $$ \dfrac{-1-2b+ab}{(1+b)^2} = -4 $$ $$ -1-2b+ab = -4(1+b)^2 $$ $$ -1-2b+ab = -4(1+2b+b^2) $$ $$ -1-2b+ab = -4-8b-4b^2 $$ $$ ab+4b^2+6b+3 = 0 \quad \text{(Equation 2)} $$ Step 5: Solve the system of equations for $a$ and $b$. From Equation 1: $a = -2-3b$. Substitute this expression for $a$ into Equation 2: $$ (-2-3b)b + 4b^2+6b+3 = 0 $$ $$ -2b-3b^2 + 4b^2+6b+3 = 0 $$ $$ b^2+4b+3 = 0 $$ Factor the quadratic equation: $$ (b+1)(b+3) = 0 $$ This gives two possible values for $b$: $b=-1$ or $b=-3$. If $b=-1$: From Equation 1, $a = -2-3(-1) = -2+3 = 1$. However, if $b=-1$, the denominator of the original curve equation becomes $(x-1)(x-2)$. Since the point $(1,-3)$ is given, $x=1$ would make the denominator zero, making the function undefined at that point unless the numerator also cancels it out. If $a=1$, the function becomes $y=\frac{x-1}{(x-1)(x-2)} = \frac{1}{x-2}$ for $x \neq 1$. For this function, at $x=1$, $y$ approaches $-1$, not $-3$. Thus, $b=-1$ is not a valid solution. If $b=-3$: From Equation 1, $a = -2-3(-3) = -2+9 = 7$. With $a=7$ and $b=-3$, the denominator $(x+b)(x-2)$ becomes $(x-3)(x-2)$. At $x=1$, the denominator is $(1-3)(1-2) = (-2)(-1) = 2 \neq 0$. This is consistent. So, the values are $a=7$ and $b=-3$. Step 6: Calculate the value of $a+b$. $$ a+b = 7 + (-3) = 4 $$ The final answer is $\boxed{4}$.