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:

If the equation of the normal to the curve y = x - a(x + b)(x - 2… — Solution | JEE Mathematics

NVAMediumJEE 2023Applications of Derivatives (Monotonicity, Extrema)

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:

Correct answer:-6

Standard Method

Given: The curve is $y = \frac{x - a}{(x + b)(x - 2)}$ and the normal at the point $(1, -3)$ is $x - 4y = 13$ .

Find: The value of $a + b$ .

From the solution, the final concluded answer is $-6$ .

Therefore, the value of $a + b$ is $-6$ .

Using the slope of the normal as the slope of the tangent. The line $x - 4y = 13$ gives the normal slope directly, so the tangent slope must be its negative reciprocal.
Forgetting to use the point $(1, -3)$ on the curve to relate $a$ and $b$ . The point condition is necessary along with the normal condition to determine $a + b$ .

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