Question

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:

Answer 1

Correct Answer: 4

(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: \[ a + b = 4. \]