ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 6. Application of derivatives

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

  1. A \(4\)
  2. B \(2\)
  3. C \(6\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(4\)

Step-by-step Solution

Detailed explanation

\(y=\frac{x-a}{(x+b)(x-2)}\) At point \((1,-3)\), \(-3 =\frac{1-9}{(1+b)(1-2)}\) \(\Rightarrow 1-a =3(1+b)\) Now, \(y=\frac{x-a}{(x+b)(x-2)}\) \(\Rightarrow \frac{d y}{d x}=\frac{(x+b)(x-2) \times(1)-(x-a)(2 x+b-2)}{(x+b)^2(x-2)^2}\) At \((1,-3)\) slope of normal is…
From JEE Mains
Explore more questions on app