Application Of Derivatives Question 4

Question 4 - 29 January - Shift 2

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________

Show Answer

Answer: 4

Solution:

Formula: Equation of normal to a curve, Slope of a normal

$ y=\frac{x-a}{(x+b)(x-2)} $

At point $(1,-3)$,

$ -3=\frac{1-a}{(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 $\frac{1}{4}$ hence $\frac{d y}{d x}=-4$,

So, $-4=\frac{(1+b)(-1)-(1-a) b}{(1+b)^{2}(-1)^{2}}$

Using equation (1)

$\Rightarrow-4=\frac{(1+b)(-1)-3(b+1) b}{(1+b)^{2}}$

$\Rightarrow-4=\frac{(-1)-3 b}{(1+b)}(b \neq-1)$

$\Rightarrow b=-3$

So, $a=7$

Hence, $a+b=7-3=4$