1. **State the problem:** We are given the curve $$y = \frac{x - a}{(x + b)(x - 2)}$$ and a point on the curve $(1, -3)$. The equation of the normal to the curve at this point is $$x - 4y = 13$$. We need to find the value of $a + b$. 2. **Find $y$ at $x=1$:** Since $(1, -3)$ lies on the curve, substitute $x=1$ and $y=-3$: $$-3 = \frac{1 - a}{(1 + b)(1 - 2)} = \frac{1 - a}{(1 + b)(-1)} = -\frac{1 - a}{1 + b}$$ Multiply both sides by $-(1 + b)$: $$3(1 + b) = 1 - a$$ So, $$1 - a = 3 + 3b$$ Rearranged: $$a + 3b = -2 \quad (1)$$ 3. **Find the slope of the normal:** The normal line is given by $$x - 4y = 13$$ Rewrite in slope-intercept form: $$-4y = 13 - x$$ $$y = \frac{x}{4} - \frac{13}{4}$$ The slope of the normal is: $$m_{normal} = \frac{1}{4}$$ 4. **Find the slope of the tangent:** The slope of the tangent $m_{tangent}$ is the negative reciprocal of the slope of the normal: $$m_{tangent} = -4$$ 5. **Find $\frac{dy}{dx}$ for the curve:** Given $$y = \frac{x - a}{(x + b)(x - 2)}$$ Let $$u = x - a, \quad v = (x + b)(x - 2) = x^2 + (b - 2)x - 2b$$ Then $$y = \frac{u}{v}$$ Using quotient rule: $$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ Calculate derivatives: $$\frac{du}{dx} = 1$$ $$\frac{dv}{dx} = 2x + (b - 2)$$ So, $$\frac{dy}{dx} = \frac{v \cdot 1 - u (2x + b - 2)}{v^2} = \frac{v - u(2x + b - 2)}{v^2}$$ 6. **Evaluate $\frac{dy}{dx}$ at $x=1$:** Calculate $u$, $v$, and $\frac{dv}{dx}$ at $x=1$: $$u = 1 - a$$ $$v = (1 + b)(1 - 2) = (1 + b)(-1) = -1 - b$$ $$\frac{dv}{dx} = 2(1) + (b - 2) = 2 + b - 2 = b$$ Substitute into derivative: $$\left. \frac{dy}{dx} \right|_{x=1} = \frac{v - u \cdot \frac{dv}{dx}}{v^2} = \frac{-1 - b - (1 - a) b}{(-1 - b)^2} = \frac{-1 - b - b + a b}{(1 + b)^2} = \frac{-1 - 2b + a b}{(1 + b)^2}$$ 7. **Set the slope equal to $m_{tangent} = -4$:** $$\frac{-1 - 2b + a b}{(1 + b)^2} = -4$$ Multiply both sides by $(1 + b)^2$: $$-1 - 2b + a b = -4 (1 + b)^2$$ Expand right side: $$(1 + b)^2 = 1 + 2b + b^2$$ So, $$-1 - 2b + a b = -4 - 8b - 4b^2$$ Bring all terms to one side: $$-1 - 2b + a b + 4 + 8b + 4b^2 = 0$$ Simplify: $$3 + 6b + a b + 4b^2 = 0 \quad (2)$$ 8. **Use equation (1) to express $a$ in terms of $b$:** $$a = -2 - 3b$$ Substitute into (2): $$3 + 6b + (-2 - 3b) b + 4b^2 = 0$$ Simplify: $$3 + 6b - 2b - 3b^2 + 4b^2 = 0$$ $$3 + 4b + b^2 = 0$$ 9. **Solve quadratic for $b$:** $$b^2 + 4b + 3 = 0$$ Factor: $$(b + 3)(b + 1) = 0$$ So, $$b = -3 \quad \text{or} \quad b = -1$$ 10. **Find corresponding $a$ values:** For $b = -3$: $$a = -2 - 3(-3) = -2 + 9 = 7$$ For $b = -1$: $$a = -2 - 3(-1) = -2 + 3 = 1$$ 11. **Find $a + b$ for each pair:** For $(a, b) = (7, -3)$: $$a + b = 7 - 3 = 4$$ For $(a, b) = (1, -1)$: $$a + b = 1 - 1 = 0$$ 12. **Check which pair satisfies the point on the curve:** Recall from step 2: $$-3 = -\frac{1 - a}{1 + b}$$ Check for $(7, -3)$: $$-3 = -\frac{1 - 7}{1 - 3} = -\frac{-6}{-2} = -3$$ True. Check for $(1, -1)$: $$-3 = -\frac{1 - 1}{1 - 1} = -\frac{0}{0}$$ Undefined, so discard. **Final answer:** $$a + b = 4$$