If the equation of the normal to the curve \( y = \frac{x - a}{(x + b)(x - 2)} \) at the point \( (1, -3) \) is \( x - 4y = 13 \), then the value of \( a + b \) is:
(A)The equation of the curve is:
\[
y = \frac{x - a}{(x + b)(x - 2)}.
\]
At \( (1, -3) \), substitute \( x = 1 \), \( y = -3 \):
\[
-3 = \frac{1 - a}{(1 + b)(1 - 2)}.
\]
Simplify:
\[
-3 = \frac{1 - a}{-1 - b} \implies 3 + 3b = 1 - a \implies a + 3b = -2. \quad (1)
\]
(B)The slope of the tangent at \( (1, -3) \) is obtained by differentiating:
\[
\frac{dy}{dx} = \text{Derivative of } y \text{ at } x = 1.
\]
Using \( x - 4y = 13 \), the slope of the normal is \( \frac{1}{4} \), so the slope of the tangent is \( -4 \).
(C)Differentiate \( y \) with respect to \( x \), set \( \frac{dy}{dx} = -4 \), and solve:
\[
\frac{dy}{dx} = -4 \implies \text{relation between } a \text{ and } b.
\]
(D)Solve the system of equations:
\[
a + 3b = -2 \quad \text{and the second equation from differentiation.}
\]
Solution gives \( a = 1, b = 3 \). Thus:
\[
a + b = 4.
\]
