Question

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

Answer 1

$y=\frac{x-a}{(x+b)(x-2)}$
At point $(1,-3)$,
$-3 =\frac{1-9}{(1+b)(1-2)} $
$\Rightarrow 1-a =3(1+b)$....(1)
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, $a+b=7-3=4$