If the equation of the normal to the curve y = (x - a)/(x + b)(x - 2) at the point (1, -3) is x - 4y = 13, then the value of a + b is equal to ___________.

Solution

Maths sir · 5 steps

Use two facts at x = 1: the point lies on the curve, and the tangent slope is -4.

1. Point on curve. Put (1, -3) into the curve: -3 = (1-a)/(1+b)(-1). This gives 1 - a = 3(1+b), that is, a + 3b = -2.
2. Tangent slope. The normal x - 4y = 13 has slope 1/4. The tangent is perpendicular to it, so y'(1) = -4.
3. Differentiate. Write y = (x-a)/D with D = (x+b)(x-2), so D' = 2x + b - 2. At x = 1 you get D = -(1+b) and D' = b. Put these into the quotient rule y' = (D - (x-a)D')/D².

   y'(1) = (-(1+b) - (1-a)b)/(1+b)² = -(1+b)(1+3b)/(1+b)² = -(1+3b)/(1+b)

4. Solve. The middle step used 1 - a = 3(1+b). Setting -(1+3b)/(1+b) = -4 gives 1 + 3b = 4 + 4b, so b = -3. Then a = -2 - 3b = 7.
5. So. a + b = 7 + (-3) = 4. Check: with a = 7 and b = -3, y(1) = -6/(-2)(-1) = -3, as required.

The idea to keep. A normal at a given point gives you two equations: the point satisfies the curve, and the tangent slope is the negative reciprocal of the normal's slope. Two unknowns, two equations.

Answer from the NTA final answer key. Solution by PrepWiser.