Question

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

Answer 1

Answer: 4

$y=\frac{x-a}{(x+b)(x-2)}$
At point $(1,-3)$,
$-3=\frac{1-9}{(1+b)(1-2)}$
$\Rightarrow 1-a=3(1+b)$....(1)
Now, $y=\frac{x-a}{(x+b)(x-2)}$
$\Rightarrow \frac{dy}{dx}=\frac{(x+b)(x-2)\times (1)-(x-a)(2x+b-2)}{(x+b{)}^2(x-2{)}^2}$
At $(1,-3)$ slope of normal is $\frac{1}{4}$ hence $\frac{dy}{dx}=-4$,
So, $-4=\frac{(1+b)(-1)-(1-a)b}{(1+b{)}^2(-1{)}^2}$
Using equation (1)
$\Rightarrow -4=\frac{(1+b)(-1)-3(b+1)b}{(1+b{)}^2}$
$\Rightarrow -4=\frac{(-1)-3b}{(1+b)}(b\ne -1)$
$\Rightarrow b=-3$
So, $a=7$
Hence, $a+b=7-3=4$